may be divided by the weight of the liquid displaced and the quotient then multiplied by the specific gravity of the liquid; by either method the specific gravity of the soluble substance will be obtained. To find the weight of a corresponding volume of water, divide the weight of the liquid displaced by its specific gravity, for the weights of equal volumes of two bodies are to each other directly proportional as their specific gravities. Example: A piece of alum weighs 125 grains in air; immersed in oil of turpentine having the specific gravity 0.860 it weighs 62 grains; 125 divided by 63 (the loss of weight) yields 1.984; oil of turpentine weighing only 0.86 as much as water, 1.984 must be multiplied by 0.860, which gives 1.7062+ as the specific gravity of the alum. Or the weight of a volume of water corresponding to the volume of oil of turpentine displaced may be found by dividing 63 by 0.86, which equals 73.256, and this divided into 125, the weight of the alum in air, also gives 1.7062+ as the specific gravity of the alum. Sometimes it is desirable to find the specific gravity of solids in powder form, as calomel, reduced iron, lead oxide, and the like; this is best done by using a flask or bottle known to hold a definite quantity of water, introducing a certain weight of the powder, and then filling with water and weighing the total contents; as two bodies cannot occupy the same space at the same time, it follows that the flask or bottle containing the powder cannot hold the same quantity of water as when empty, and this difference corresponds to the weight of water equal in volume to the powder. Suppose 100 grains of an insoluble powder are placed in a counterpoised bottle capable of holding exactly 1000 grains of water, the latter being then filled with pure water; if the total contents weigh 1088 grains, 12 grains of water have been displaced by the powder, for 1088 — 100 leaves 988, and as the bottle is capable of holding 1000 grains of water, the difference 1000-988-12 must have been displaced. Then applying the rule, 8.333 is found to be the specific gravity of the powder, as 100 = 12 8.333 . SPECIFIC VOLUME. The term specific volume is used to define the ratio existing between the volumes of certain weights of bodies and the volume of the same weight of pure water; it is therefore the opposite of specific gravity. Specific volume is ascertained by dividing the specific gravity of a body into unity, and hence may be called the reciprocal of specific gravity; it may also be found by dividing the weight of a given volume of water by the weight of an equal volume of a liquid. Every pharmacist is aware that it will require vessels of different size to hold one pound of ether, water, glycerin, sulphuric acid, oil of turpentine, or chloroform, and it is often desirable to know in advance the volume of a given weight of a liquid. In the metric system this is a very simple operation, for the weight in grammes of any liquid multiplied by the specific volume, or divided by the specific gravity, of that liquid at once expresses the actual volume in cubic centimeters. To find, however, the volume of a given weight, avoirdupois or apothecaries', of a liquid, it becomes necessary first to ascertain the volume of a like weight of water, and then to multiply this by the specific volume, or to divide by the specific gravity of the liquid; or the given weight of a liquid may be divided at once by its specific gravity, which will yield the weight of a volume of water equal to the volume of the liquid, and then by finding the volume of such a weight of water the volume of the liquid is at once known. Examples: If the volume of 500 Gm. of alcohol U. S. P. is desired, divide 500 by 0.820, the specific gravity of the alcohol, and the quotient 609.75+ will be the answer in cubic centimeters. To find the volume of 8 ounces of official glycerin (apothecaries' weight) it is necessary to multiply by 480, the number of grains in 1 ounce, and then divide the product by 455.7, the number of grains in one U. S. fluidounce of water, the quotient (480 X 83840; 3840455.78.427), 8.427, represents the number of fluidounces contained in the same weight of water; 8.427 then divided by 1.25, the specific gravity of the glycerin, yields 6.7416 fluidounces as the volume of 8 ounces, apothecaries' weight, of glycerin. How large a bottle is required to hold 1 pound of chloroform of 1.490 specific gravity? One pound avoirdupois is equal to 7000 grains, and 7000 1.490 4697.986, the weight in grains of a volume of water equal to the chloroform; then 4697.986 ÷ 455.7 10.309, or very nearly 10 fluidounces. How many fluidounces in 2 pounds of official ether having a specific gravity of 0.725. Two pounds of water measure 30.72 fluidounces, and multiplying this by the specific volume, 1.379, of the ether (1.000 0.725 1.379+) we obtain 42.36 fluidounces as the answer. ADJUSTMENT OF SPECIFIC GRAVITY AND PER CENTAGES. While the adjustment of percentages in liquids as well as solids presents no difficulties, the reduction of liquids from a higher to a lower specific gravity is not quite so easily accomplished, since specific gravity is but the expression of the relation between volume and weight, and condensation of volume generally occurs as the result of a mixture of two liquids. Two very simple rules, or formulas, have been published for the adjustment of specific gravities of liquids, by volume and by weight; but absolutely accurate results are only possible when no contraction of volume takes place; in the majority of cases the condensation of volume is but very slight, and for ordinary purposes may be ignored. It is well known that the weights of equal volumes of two liquids are to each other directly proportional as the specific gravities of these liquids ; there fore the weight of a liquid divided by its specific gravity represents a weight of water equal in volume to that liquid. It is also well known that the volumes of equal weights of two liquids are to each other inversely proportional as the specific gravities of these liquids; therefore, the volume of a liquid multiplied by its specific gravity represents a volume of water equal in weight to that liquid. The well-known process of alligation is admirably adapted to the adjustment of specific gravities of liquids by volume, but is unsuited to adjustment by weight. When two liquids of different specific gravities are mixed, the loss which one suffers will be balanced by the gain of the other; hence the two liquids used must be mixed in inverse proportion to that existing between the gain and loss of specific gravity and the specific gravity of the mixture; the difference between the higher specific gravity and the desired specific gravity of the mixture will therefore indicate the proportion of the liquid having the lower specific gravity; and the difference between the lower specific gravity and the desired specific gravity will indicate the proportion of the liquid having the higher specific gravity. For example, if solution of ferric chloride, specific gravity 1.520, is to be reduced to 1.387 specific gravity by addition of a weaker solution of 1.280 specific gravity, 107 volumes of the stronger must be mixed with 133 volumes of the weaker solution; or, in other words, 1 volume of the former with 1.243 volumes of the latter. It is customary to set down a problem in alligation in the following manner to facilitate comparison : 1.387 { 1.520 0.107 1.280 | 0.133 proportion of the stronger liquid. proportion of the weaker liquid. If a definite volume of the mixture is desired, the requisite volume of the stronger and weaker liquids may be ascertained by dividing the desired volume by the sum of the proportionals, and then multiplying each proportional by the quotient so obtained; thus, if 32 fluidounces are wanted, divide 32 by 0.240 (0.107 +0.133), which yields 133.3; 0.107 × 133.3 = 14.27 fluidounces, the requisite volume of the stronger solution, and 0.133 × 133.3 = 17.73 fluidounces, the requisite volume of the weaker solution. To adjust the specific gravity of a given weight of a liquid to a higher or lower specific gravity, the following formula may be employed: in which represents the weight of the diluent, w the weight of the liquid to be diluted, a the specific gravity of the liquid to be diluted, the desired specific gravity, and e the specific gravity of the diluent. (Whenever water is the diluent, e is made 1.000.) As stated before (see above), W weight of water equal in volume to w, a Example: How much water must be added to 250 Gm. of solution of potassa of 1.539 specific gravity in order to reduce the specific gravity to 1.036? Substituting numerical values for the letters 250 X 1.000 (1.539 1.036) in the above formula, we have a 1.539 (1.036 - 1.000) 2269.6. Answer: 2269.6 Gm. then 250 X 0.503 1.539 × 0.036 125.75 0.055404 To make a definite weight of a liquid of definite specific gravity by mixing two liquids of known specific gravity, both being of the same kind, or one being water: Let me represent the desired weight of the mixture, a the weight of the diluent, y the weight of the liquid to be diluted, and a, b, c the specific gravity of the liquid to be diluted, of the mixture desired, and of the diluent, respectively. Since x+y=mw, and the value of r has been shown above to be the weight of the liquid to be diluted ×_c (a − b), a (b − e) the latter expression may be substituted for a in the equation, x + y y× =mw; thus c(ab) +y: mw. This simplified is yea―yeb +yabyacmwa (bc), and cancelling, yb (ac) × a i (b − e). a (b — c) = mwX ; mwa (b-c) The value of y (weight of stronger liquid) having been ascertained, it is subtracted from mr, the desired weight of the mixture, to find the value of r, the weight of the diluent. Example: If it is desired to make 10 pounds of ammonia-water of 0.960 specific gravity, from ammonia-water of 0.900 specific gravity, mix 3.75 pounds of the latter with 6.25 pounds of water; for, substituting numerical values for the letters in the above formula, the weight of the liquid to be diluted is equal to 10X0.900 (0.960- 1.000) 10 X -0.036 - 0.36 -0.096 3.75, and This example may also be worked out by alligation, which method would give the proportional parts of the volume to be used of each liquid, as follows: 1.000 0.060 or 6 0.9600.900 0.040 or 4 But as weight and not volume is called for, it becomes necessary to find the actual weights of the respective volumes by multiplying the latter by the specific gravities of the liquids; thus, The result shows that the mixture would produce only 9.6 parts by weight, and since 10 parts (or pounds) are wanted, the respective necessary quantities may be found by the rule of three; thus, For the adjustment of percentage in alcohol (by weight or volume), in acids (by weight), and in alkali solutions (by weight), the following rules may be applied: For reducing solutions from a higher to a lower percentage: Multiply the given quantity by the given percentage and divide by the required percentage; the quotient will be the quantity to which the liquid must be diluted by the addition of water. Since alcohol is frequently reduced in volume percentage, and contraction of volume invariably follows the admixture of alcohol and water, it becomes necessary, after contraction has ceased, to add sufficient water to restore the original volume of the mixture. Examples: Reduce 4 pints (64 fluidounces) of 93 per cent. (by volume) alcohol to 65 per cent.: 64 × 93 5952, and 5952 ÷ 65 =91.57. Enough water must be added to the 4 pints of alcohol to yield, after contraction has ceased, 91.57 fluidounces. × Reduce 2 pounds of hydrochloric acid from 31.9 per cent. to 10 per cent. 2 pounds 32 avoirdupois ounces; 32 X 31.9 = 1020.8, and 1020.8 10: 102.08. Enough water must be added to the 2 pounds of acid to bring the total weight up to 102.08 avoirdupois ounces. Reduce 8 troy ounces of stronger ammonia-water, 28 per cent., to 10 per cent. strength: 8 X 28=224, and 224 ÷ 10-22.4. Enough water must be added to the 8 troy ounces of stronger ammonia-water to bring the total weight up to 22.4 troy ounces. For making a definite quantity of a solution of a certain percentage by diluting a stronger solution with water: Multiply the required quantity by the required percentage, and divide by the higher percentage; the quotient will be the quantity of the stronger liquid necessary, and this subtracted from the total quantity required leaves the necessary quantity of water. When volume adjustment of alcohol is made, the |