Materia medica and therapeutics, a text-book for nurses

Preface

The aim of this book is to present the use of drugs from a scientific basis in such a way as to appeal to the nurse's interest. An effort has been made to give only the important and practical points which form a foundation for an intelligent handling of drugs, but not for prescribing them. The classification by systems seems the most practical from the nurse's point of view, and it also gives an opportunity for the review of Anatomy and Physiology. In Toxicology only the striking symptoms of poisoning are given, because they can be more easily remembered than detailed descriptions. First aid treatments are emphasized.

The study of Materia Medica needs to be supplemented by experiments, demonstrations, and observations in the wards. The number of experiments suitable for hospital class work is very limited. In the discussion of any drug it is important that the forms in which it is used as a medicine should be shown before the class unless they are very well known. Detailed descriptions of the physical properties of drugs have in many cases been omitted, as familiarity with their appearance can only be gained by actually seeing them.

iv PREFACE

A nurse seldom handles crude vegetable drugs, but a list is given (page 87) of a few which will add interest to the lesson.

An effort has been made to give a broader interest in this field than is obtained in the hospital and the sick room. A nurse should have some conception of the world-wide drug industry that she may appreciate the work of the many trained scientific men engaged in it, and on the other hand, that she may be cognizant of the evils of patent medicines and nostrums.

The author acknowledges her indebtedness to Dr. Cary Eggleston, instructor of pharmacology in Cornell Medical School, New York City, for his criticism of the manuscript and for many valuable suggestions.

L. A. P.

New York, 1915. •

PHARMACY,

Chapter I. Weights And Measures

In the United States, unfortunately, there are in use several systems of weights and measures in handling drugs, the apothecaries', avoirdupois, household, and metric. The apothecaries' and avoirdupois systems, which have their origin and legal standards in England, have been in most common use, but are gradually being replaced by the metric system, the French system of measure. Household measures, cups, spoons, and the like, frequently have to be used in the home.

In the work-room of every pharmacy there are sets of apothecaries' and of metric weights and measures. These are used in measuring drugs for prescriptions and in making up stock preparations, such as liniments, spirits, and solutions. On the outside counter of the store, however, all drugs sold in bulk are weighed according to the avoirdupois system. 2

METRIC SYSTEM.

Table of Weights.

1 gramme (Gm. 1 ) = 10 decigrammes (dg.) = 100 centigrammes (eg.) = 1000 milligrammes (mg.) 10 grammes = 1 decagramme (Dg.)

100 grammes = 1 hectogramme (Hg.)

1000 grammes = 1 kilogramme (Kg.)

The decagramme and hectogramme are not much used, large quantities being expressed as 15 Gm. instead of 1 Dg. and 5 Gm. The ending -me, especially in gramme, should always be used in spelling these words, because, in writing, the words grain and gram are easily confused. The prefixes of the denominations below a gramme are duplicated in our system of money, and a comparison will help in remembering the names and their relation.

1 dollar = 10 dimes 1 gramme = 10 decigrammes = 100 cents = 100 centigrammes

= 1000 mills = 1000 milligrammes

The table of weights below one gramme may be written as follows:

1 gramme = 1 . 000 gramme

1 decigramme = 0.1 1 centigramme = 0.01 1 milligramme = 0.001 It is then seen that each decimal place in the gramme represents one of the lower denominations, e. g., the

1 The abbreviations used in this text for gramme, Gm., and for cubic centimeter, Cc, are those used in the United States Pharmacopeia. The framers of the pharmocopeia adopted the form Gm. in preference to gm. to avoid confusion with gr., the abbreviation for grain, and the form Cc, instead of cc, to make the two metric abbreviations uniform.

METRIC SYSTEM 19

first place is taken by decigrammes as they are one-tenth of a gramme and 2.3 grammes may be read 2 grammes and three decigrammes or 23 decigrammes. 4.25 grammes may be read 4 grammes, 2 decigrammes, and 5 centigrammes, or 425 centigrammes. Therefore, converting a figure in the metric system from one denomination to another is simply a matter of moving the decimal point, which is equivalent to multiplying or dividing by 10, 100, or 1000. Move it to the left if converting to a larger denomination, to the right if to a smaller. Move it one place if the denomination is larger or smaller by 10, two places if by 100, and three places if by 1000.

If it is known that a quantity is expressed in the metric system the denominations are often understood without being written. 4.5 in the metric system is always grammes, 0.008 is eight thousandths of a gramme, 0.8 is eight tenths of a gramme. The denominations, centigramme and decigramme are not commonly used, a quantity being expressed as 500 mg. or 0.5 gm. instead of 5 dg., and 20 mg. or 0.02 gm. instead of 2 eg.

Problems.

1. Convert the following to grammes by changing the decimal point:

2. Convert the following to grammes:

3. Convert the following to milligrammes:

4. Add the following and express the sum in grammes:

4.620

5. Add and express sum in grammes:

6. Add and express sum in milligrammes:

7. The dose of morphine sulphate is 15 mg. How many doses could be given from one gramme?

8. The dose of strychnine sulphate is . 001 Gm. How many mg. in 4 doses?

9. The dose of hexamethylenamine (urotropin) is 250 mg. How many grammes would you need for 4 doses?

10. The dose of aspirin is 3 dg. How many doses in three grammes?

Tables of Volume and Length. — These are on the same basis as the table of weights, and can be easily written out from that table by substituting in the names of the different measures liter for gramme for the table of volume and meter for gramme for the table of length. For example, the table of volume is 1 liter = 10 deciliters, 100 centiliters, and so on; the table of length 1 meter = 10 decimeters, 100 centimeters, and so on. Pharmacists are sometimes called upon to use linear measure in making plasters of prescribed size; a nurse probably would never use it. In the table of volume the liter and cubic centimeter are practically the only measures in common use. 1 liter equals 1000 milliliters and 1 milliliter is a quantity of water which at 4° C. would occupy 1 cubic centimeter of space. Therefore, the term milliliter is synonomous with that of cubic centimeter and the latter term is generally used. The common equivalents are as follows:

1000 cubic centimeters (Cc.) = 1 liter (L.). 1 Cc. at 4° C. weighs 1 gramme

Therefore 1000 Cc. or 1 L. at 4° C. weighs 1000 Gm. or 1 Kg.

APOTHECARIES' SYSTEM 21

Problems.

1. How many Cc. of water would you measure out to get the following:

2. How much would the following quantities weigh:

3. How many Cc. in 2.6 L.?

4. The dose of atropine sulphate is 0.4 mg. What fraction of a gramme is needed to make 30 Cc of a solution in which 1 Cc.. would contain the dose?

5. The dose of tincture of opium is 0.5 Cc. 10 Cc. of the tincture contain 1 Gm. of opium. 12 per cent, of opium is morphine. How many milligrammes of morphine in one dose of the tincture?

APOTHECARIES' SYSTEM.

Table of Weights.

20 grains (gr.) = 1. scruple O) 1 3 scruples = 1 dram (5 ) = 60 grains

12 ounces = 1 pound (lb.) = 5760 "

The scruple is seldom used, all quantities less than a dram being expressed in grains. The quantity \ is expressed by the symbol ss. When the symbols are used the quantity is expressed in Roman numerals and written after the symbol, e. g., 2 grains is written gr. ij; 5 drams, 5v; a half of a grain, gr. ss.

1 About the time of Christ signs were in common use to express quantities. The apothecaries' signs 9, 3, 5, used at the present time, are believed to have originated from these, and they differ only slightly from the ones which in those early times designated about the same quantities. The abbreviation for pint (O) and for gallon (C) (see table of liquid measure) are derived from the Latin words congius, a gallon, and octarius, one-eighth of a gallon.

Problems.

1. How many grains in §iv; 3v; Biiss?

2. Salt §j will make how many quarts of saline solution (gr. xc to 1 qt.)?

3. One Seidlitz powder contains sodium bicarbonate gr. xl. How many drams of sodium bicarbonate in six powders?

4. How many drams of salt would you need to make 4 qt. of saline solution (gr. xc to 1 qt.)?

5. How many ounces of mucoi are needed to make 2 qt. of solution (gr. lx to Oj)?

Table of Liquid Measure.

60 minims (TTl) = 1 fluidram (f5)

8 fluidrams = 1 fluidounce (f § ) = 480 minims 16 fluidounces = 1 pint (O) = 7,680 "

4 quarts = 1 gallon (C)

These symbols and quantities are written like those of weights: 7 minims = ITlvij, 5 fluidounces = f§v. In common practice the symbols for a liquid ounce and dram are the same as for the same quantities by weights, it being understood that the substance is a liquid. For example, 5 fluidounces of alcohol are expressed §v.

Problems.

1. The dose of a solution is Tfl,xxx. This dose contains gr. -§■$ of strychnine. How much of the drug in f S -1 of the solution?

2. A solution of hexamethylenamine is labelled f^ss = gr. vj. How many drams of the solution would you use to give gr. ix of the drug?

3. Morphine gr. vj is contained in fgj of a solution. How much morphine in Tfl,x?

4. How many ounces of salt solution (gr. xc to 1 qt.) could be made with gr. xxx of salt?

5. 3ij of a solution contain gr. j of cocaine. How much cocaine is given when a doctor orders TTlx of the solution?

HOUSEHOLD MEASURES 23

AVOIRDUPOIS SYSTEM OF WEIGHTS.

437| grains = 1 ounce (oz.) 16 ounces = 1 pound (lb.) = 7000 gr. Two points of difference should be noted between this and the apothecary system of weights: (1) the smaller number of grains in the ounce, and (2) the greater number of ounces in the pound.

Problems.

(The weighing in the following is done by apothecaries' weights; the drugs bought in bulk are weighed according to the avoirdupois system.)

1. A pound of salt will make how many quarts of saline (gr. xc to 1 qt.)?

2. Four ounces of soda bicarbonate will make how many Seidlitz powders (gr. xl in each powder)?

3. If the pharmacy nurse buys 1 oz. of trional how many powders of gr. x can she make?

HOUSEHOLD MEASURES.

60 drops (gtt.) = 1 teaspoonful (t.).

4 teaspoonfuls = 1 tablespoonful (T).

2 tablespoonfuls = 1 fluidounce

2 fluidounces = 1 wineglass.

6 fluidounces = 1 teacup.

8 fluidounces = 1 glassful.

Drops, teaspoons, tablespoons, and cups vary so much in size that this system should never be used if it can be avoided. A drop of water at ordinary temperature is considered equivalent to a minim, and both approximately weigh 1 grain/ The difference between a minim and a drop can be easily demonstrated by expelling 10 minims of different fluids, drop by drop, from a graduated minim dropper. It will be seen that 10 minims make about 20 drops of water, about 40 drops of alcohol, and less than 20 drops of heavy oils. Therefore, if a doctor orders 10 minims of an alcoholic preparation, like a tincture, and 10 drops are given, the patient gets only a portion of the dose.

Pkoblems.

1. State how you would measure quantities for the following orders using household measures: (a) Douche, 3J to a pint. (6) Injection §ij to §iv.

(c) Nutritive enema:

Malted milk, gss

Somatose, 3J

Water, § i v

Sodium bicarb., gr. xx

(d) Enema:

White of egg

Glycerin, §ij

Oil of turpentine, 5ij

Water, Oj

APPROXIMATE EQUIVALENTS.

1 gramme

= gr. xv.

1 cubic centimeter

=

TUxv.

1 grain

=

1 fluidram

=

4Cc.

1 dram

=

4Gm.

1 fluidounce

=

1 ounce

=

1 quart

=

IL.

USE AND CARE OF MEASURING APPARATUS 25

The equivalents for fractions of a grain are obtained by considering 1 grain equal to 60 milligrammes. Then gi^ of a grain equals 1 milligramme and the other fractions in common use can be easily reckoned. Any measure under 15 grains or minims is expressed in a decimal fraction of a gramme or cubic centimeter. The approximate equivalents in common use from 1 to 15 grains are expressed in round numbers as in the table below. Those for minims in cubic centimeters are the same as grains in terms of grammes.

1 grain = 0.064 gramme. 5 grains =0.3 gramme.

2 " =0.1 " 7, 7|, 8 " =0.5

3 " =0.2 " 10 " =0.6

4 " =0.25 " 15 " =1.0

Problems.

1. Express the following in grains:

2. Give the equivalents in the apothecaries' system:

3. If only a Cc. measure were available how would you obtain the following:

Oj; f§v; 1 T.; f3iij; a glassful; \ L.; 10 Gm.?

4. With household measures how would you obtain the following:

3iv; 2 Gm.; fgss; mxx; 1 L.; 20 Cc: §j?

5. With metric weights how would you weigh the following:

gr. xxx; §ss; 5ij; gr. -^ ?

USE AND CARE OF MEASURING APPARATUS.

All weights should be kept in a closed box. An accumulation of dust will make them inaccurate, especially the small ones. Small weights should be handled with forceps only. No weights should be so scrupulously cleaned as to polish off the lacquer or metal and thus reduce their weight. When not in use balances ought to be covered, and at all times kept clean.

Balances are gauged to weigh quantities within certain limits, as from 0.1 to 10 Gm., and are not accurate beyond those limits. Before using a balance make sure that it is in equilibrium, and that it is not in a draught of air. The weights should be placed on the

Fig. 1. — Cup on surface of liquid. A, line at which reading is taken.

pan gently, as good balances are delicately made and a sudden jar might injure the mechanism.

All glass measures (no others should ever be used) must be held on a line with the eyes while being read. The surface of most liquids in a cylindrical vessel is cupped, as can be easily seen in a test-tube (see Fig. 1). All glass measures are standardized to be read at the base of this cup, and unless so read the quantities are not correctly measured.

Chapter II. Solutions

True Solution. — A true solution is a clear, transparent, homogeneous liquid having a gas, liquid, or solid dissolved in it. Hydrogen peroxide is an example of a true solution of a gas; solutions of alcohol and of salt are other examples of true solutions. There are some substances which may be thoroughly mixed with a liquid and the resulting mixture appear homogeneous temporarily, but after a time the ingredients separate. Such mixtures are called suspensions. All emulsions are suspensions, as is also a common preparation called Compound Chalk Mixture, which is used for diarrhea. (See Experiment I.)

Parts of a Solution.— There are two parts in every solution, the liquid which forms the bulk of the solution and the substance dissolved in it. The liquid is called the solvent; the dissolved substance the solute. The most common solvent is water. Alcohol is often used, for example in the spirit of camphor and other spirits. Ether is much used in scientific laboratories and is one of the solvents in collodion. The dissolved substance, as stated above, may be a gas, a liquid, or a solid.

Solubility. — In science no substance is considered insoluble, but some are much more easily soluble than others. The solubility of a substance varies with the solvent used. Oils, for example, are very slightly soluble in water and completely soluble in alcohol. A very important and practical factor which affects the solubility of a substance is the temperature of the solution when made. Most substances are more soluble in a hot solvent than in a cold one, and for this reason a given amount of boric acid, for example, can be dissolved more quickly in hot water than in cold. (See Experiment II.)

Saturation. — When a solvent has taken up so much of a substance that any more added simply settles to the bottom, the solution is said to be a saturated solution at that temperature. The following table shows the approximate quantity required to make a saturated solution of some of the common substances.

Drug.

Apoth. weight.

Metric weight.

Per cent.

Boric acid .

6 drams = 1 pt.

Borax (sodium borate)

6 drams = 1 pt.

Sodium bicarbonate

10 drams = 1 pt.

Sodium chloride

4 ounces = 1 pt.

Calcium oxide

1\ grains = 1 pt.

0.1

Sodium phosphate

2 ounces = 1 pt.

Magnesium sulphate

8j ounces = 1 pt.

Method of Expressing Strength. — The strength of a solution may be expressed in two ways: by a (1) ratio or by a (2) percentage. The term per cent, means parts per 100. Therefore, a solution of alcohol 5 per cent, in strength means that in 100 equal parts of the solution 5 are alcohol. A ratio means that in the number of parts expressed in the second term one part is the solute. That is, in a 1 to 20 solution of alcohol one part in every 20 is alcohol. The per cent.

RECKONING THE PERCENTAGE STRENGTH 29

and ratio do not represent definite quantities, such as ounces or pints, but express a relation between the solute and the whole solution. That is in any quantity of a 5 per cent, solution of alcohol the amount of alcohol is in the same relation to the whole amount of the solution as 5 is to 100. Also in a 1 to 20 solution the relation between solute and the whole solution is the same as 1 is to 20. For example in 40 Cc. of a 5 per cent, solution there are 2 Cc. of solute because 2 is in the same relation to 40 as 5 is to 100. In 16 ounces of a 1 to 4 solution there are 4 ounces of solute, because 4 is in the same relation to 16 as 1 is to 4. As a per cent, or ratio assumes that the solution is divided into equal parts the quantity of the solute and of the whole solution must be expressed in the same denomination. For example, in 100 Cc. of a 5 per cent, solution of alcohol 5 Cc. are alcohol; in 100 minims, 5 minims are alcohol; in 20 ounces, 1 ounce is alcohol.

Method of Reckoning the Percentage Strength.

The percentage strength of a solution may be reckoned in two ways: (1) by forming a proportion, or (2) by forming a fraction and taking that fraction of 100.

Problem. — What is the percentage strength of a solution 30 Cc. of which contain 6 Cc. of solute? First Method: 6 : 30 : : x : 100. 30x = 600 x = 20

Ans. 20 per cent.

Second Method: Reasoning that as 6 is ^ of 30 the per cent, must be \ of 100; the actual figuring is -£$ or t of 100 = 20 per cent., Ans.

Rule. — 1. Form a proportion, letting x = the per cent, or 2. Multiply 100 by the fraction: amount of solute, amount of the whole solution.

Problems.

1. If 100 Cc. of a solution contain 40 Cc. of peroxide what per cent, solution is it?

2. What percentage strength are the following:

(a) 40 Cc. of solution contain 4 Cc. of peroxide.

(b) 90 Cc. of solution contain 3 Cc. of phenol.

(c) 20 Cc. of dilute alcohol contain 4 Cc. of alcohol.

(d) f§ij of dilute alcohol contain 5iij of alcohol.

(e) f5hj of dilute alcohol contain Tfl,xlv of alcohol. (/) f3J of a solution contains ITlxij of phenol.

Method of Reckoning the Strength by a Ratio.

When the strength of a solution is expressed in a ratio the amount of the solute is compared to 1 as a fixed basis. That is, if 20 Cc. of a solution contain 2 Cc. of alcohol it is said to be a 1 :10 solution; because if the 2 Cc. be considered as a unit there would be 9 other such units (18 Cc), or 10 in all. The first term of a ratio is always 1 and the second term has to be determined. It may be done in two ways:

1. By forming a proportion: 2 : 20 : : / : x

x = 10, hence the ratio is 1 : 10.

2. . By reasoning that if 1 is \ of the quantity of the solute, the second term must be \ of the amount of the

METHOD OF CONVERTING PER CENT. TO RATIO 31

whole solution. Then \ of 20 = 10 and the ratio is

1:10.

Rule. — 1. Form a proportion letting the x = the

second term of the ratio.

2. Multiply the amount of the whole solution by the fraction,

amount of solute.

Problems.

Express the strength of the following in a ratio:

100 Cc. of solution contain 20 Cc. of alcohol. 80 ounces 40 ounces

60 drams " 10 ounces

6 pints " 2 pints

fgij " fgiv

1 liter " 200 Cc.

f3ij " mxi

f3i " mij

o j " mxc

Method of Converting Per cent, to Ratio, and Vice Versa.

Sometimes the strength of a solution is expressed in percentage and one wishes to express it in a ratio, or vice versa. This may be done by forming a proportion in each case, letting x represent the per cent, or the second term of the ratio, as the case may be, or by the following method, which is simpler:

To express 4 per cent, in a ratio.

As the first term of the desired ratio is always 1, the problem is to find the second term.

4 per cent. = 4 parts in 100. or =4: 100.

The desired ratio = 1 : x x or the second term = \ of 100 = 25.

Rule. — Take 1 as the first term and divide 100 by the per cent, to obtain the second term. To express 1 : 20 in a per cent.

As per cent, means so many parts in 100, the per cent, would be as many times 1 as 100 is times 20 or 5 per cent.

Rule. — Divide 100 by the second term of the ratio.

Problems.

1. Some solution bottles are labelled as follows. State how you would label them by ratios:

10 per cent.; 1 per cent.; 33 per cent.; 0.1 per cent.; 5 per cent.; 0.5 per cent.

2. State how you would mark the following labels that the percentage strength may also be expressed:

Relative Strength of Solutions.

Many nurses have difficulty in determining the comparative strengths of solutions, that is, whether a 5 per cent, solution is stronger or weaker than a 2 per cent. The principle of this problem is the same as in interest on money, the larger the percentage of interest on an investment the greater is the income from it. So in solutions the larger the per cent, the stronger the solution.

Problems.

1. If a series of bottles of solutions were marked as follows, how would you arrange them in order of their strengths?

5 per cent.; 20 per cent.; 2 per cent.; 3 per cent.: 0.1 per cent.

2. Would it be possible to make the following solutions?

(a) A 6 per cent, from a 2 per cent. (6) A 10 per cent, from a 20 per cent.

(c) A 50 per cent, from a 20 per cent.

(d) A 50 per cent, from a 100 per cent.

METHODS OF MAKING SOLUTIONS 33

If the strengths are expressed in ratios they vary according to the second term. It is easily seen that a solution which contains 1 Cc. of alcohol in 5 of water is stronger than one which contains 1 in 20. Therefore the strengths of solutions expressed in ratios vary inversely as the second term, i. e., the smaller the second term the stronger the solution.

Problems.

Arrange solution bottles marked as follows according to their strength :

METHODS OF MAKING SOLUTIONS.

In pharmacy all solutions are made according to weight. As the kind of liquid and its temperature affect the weight of a given volume, the weights have to be reckoned according to specific gravity and temperature. Such accuracy in hospital work is unnecessary, and the weight of all liquids is considered the same as water at 4° C. That is, 4 Cc. of water, alcohol, or liquid phenol is assumed to weight 4 Gm. Therefore, if 1 ounce of alcohol is added to 99 ounces of water (a 1 per cent, or 1: 100 solution) the solution will measure 100 ounces and weigh 100 ounces. If, however, 1 ounce by weight of salt is added to 99 ounces of water the salt will dissolve and not increase the volume of water. In this case the finished solution will weigh 100 ounces, but will measure only 99 fluid- 3 ounces. It is, however, a 1 per cent, solution. There can be no uniformity in making solutions unless this principle is followed in practice.

In regard to solutions the most common problems for the nurse are to make a certain amount of a solution from crude material and to make weak solutions from strong ones kept in stock.

1. Solutions from Crude Materials.

Problem. — How much alcohol is required to make 10 ounces of a 20 per cent. (1:5) solution?

First Method. — Make a proportion letting x = amount of alcohol.

Second Method. — (a) Strength expressed in per cent. 20 per cent, means that to°o of the whole solution is alcohol. Hence, to% or i of 10 = 2 ounces.

(b) Strength expressed in ratio. This means that 1 part in 5 or i of the whole is alcohol, hence i of 10 = 2 ounces.

RULE. — First Method. — Form a proportion, letting x represent the quantity of crude material.

Second Method (a). — Strength expressed in per cent.

given per cent.

Multiply the amount of solution by the fraction,

(6) Strength expressed in ratio. Form a fraction

of the ratio and multiply the amount of solution by it.

METHODS OF MAKING SOLUTIONS 35

Problems.

(It must be remembered that the answer is in the same denomination as that in which the whole solution is expressed. It is often desirable to reduce the latter to a smaller denomination before working the problem.)

1. How much liquid phenol would be needed to make the following solutions:

(a) 40 Cc. of a 10 per cent, solution.

(b) Oiv of a 25 per cent, solution.

(c) 1 gallon (reduce to ounces) of a 1 : 20 solution.

(d) 1 liter (reduce to Cc.) of a 1 : 40 solution.

(e) 1 quart (reduce to drams) of a 20 per cent, solution.

2. Calculate amounts needed in the following solutions:

(a) 1 pint of 1 per cent, saline. (6) 1 quart of 1 : 18 boric acid.

(c) §j of 2 per cent, cocaine.

(d) 30 Cc. of 1:6 glycerin.

(e) 1 gallon of bichloride of mercury 1:1000 using 1\ gr.

tablets.

2. Weak Solutions from Strong Ones.

To understand the principles upon which these solutions are made, it must be kept in mind that the finished solution is made up of a certain number of equal parts, one of which is the original stock solution. That is, if one wishes to make a solution one-half as strong as stock he takes a certain amount of the stock and adds an equal amount of water to it. Then the finished solution is composed to two equal parts, one of which is stock. There may be 2 pints, 1 of stock and 1 of water, or 6 ounces, 3 of stock and 3 of water, but the two parts are equal. The number of equal parts is obtained by dividing the larger per cent, by the smaller. For example, to make 1 pint of a 60 per cent, solution into a 20 per cent., that is, one-third as strong, it must be diluted twice by adding two pints of water. The finished solution then will contain 3 equal parts (in this case pints), as many as 20 is contained times in 60. If a definite quantity of the weak solution is wanted, such as 12 pints of a 20 per cent, solution from a 60 per cent., the .reasoning is that as there will be three equal parts in the 12 pints, one of which is stock, there must be needed | of 12 or 4 pints of stock to make 12 pints of a 20 per cent, solution. If the strength of the solution is expressed by a ratio the second terms are divided to obtain the number of equal parts.