solid indicates the volume of water displaced, and its corresponding weight is readily noted. Suppose a solid body weighing 7.5 Gm., placed into 40 Cc. of water, causes the latter to rise to 41.5 Cc., showing that 1.5 Cc. of water have been displaced, which weigh 1.5 Gm.; then, applying the rule, 7.5 = 1.5=5, the specific gravity of the solid. Since solid bodies will float indifferently in any liquid having the same specific gravity as themselves, advantage may be taken of this property to determine the specific gravity of solids. Hager recommends determining the specific gravity of fats by placing them in alcohol and then adding water until the fat floats about indifferently beneath the surface of the mixture; the specific gravity of the mixture is then taken in the usual way, preferably by means of a pycnometer, and this at the same time expresses the specific gravity of the solid. To ascertain the specific gravity of solids insoluble in, but lighter than water, it becomes necessary to insure their immersion in water by attaching to them some heavy substance, the weight of which in water must previously have been ascertained. Upon immersing the two bodies in water it will be observed that the weight of the two appears less than the weight of the heavy body alone, which is due to the fact that the volume of water equal to the volume of the lighter body is heavier than the latter, and therefore exerts a greater upward pressure on the heavy body, causing it to appear to lose weight. The difference between the weight of the heavy body in water and the united weight of the light and heavy bodies in water expresses the excess of weight of a volume of water over the weight of a like volume of the light body; in other words, it shows how much heavier a volume of water is than the same volume of the light body; to find the exact weight of a volume of water equal to the volume of the light body, this difference, or excess, must be added to the weight of the light body in air. Suppose a piece of cork weighs 62.5 grains in air; attached to a piece of metal which weighs 94 grains in water, the whole is found upon immersion in water to weigh 88 grains, or 6 grains less than the metal alone; adding 6 to 62 5 grains (the weight of the cork) we obtain 68.5 grains, the weight of the water displaced by the cork. The specific gravity of the cork is found by dividing 62.5 by 68.5 according to the general rule on page 46. The answer will be 0.9124+. For solids soluble in water some other liquid must be selected for immersion, in which the solid body is perfectly insoluble and of which the specific gravity is known; in other respects any of the preceding methods may be followed. In such cases the weight of the liquid displaced, having been ascertained, may be used to find the weight of a corresponding volume of water, and the latter then be divided into the weight of the solid; or the weight of the solid in air 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 1000-grain bottle, 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; 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 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, avd 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 8=3840; 3840 = 455.7= 8.427), 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 troy ounces of glycerin. How large a bottle is required to hold one pouud 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 103 fluidounces. Adjustment of Specific Gravity and Percentages. 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 frequently 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 ; therefore, the weight of a liquid divided by its specific gravity represents a weight of water equal in volume to the weight of 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 the volume of 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 : 1.520 | 0.107 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 X 133.3=14.27 fluidounces, the requisite volume of the stronger solution, and 0.133 x 1333= 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 : Nr X c (a - b) c) in which a 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, b the desired specific gravity, and c the specific gravity of the diluent. (Whenever water is the diluent, c is made 1.000.) As stated before (see page 63), ":= weight of water equal in volume to w, weight w + x of water equal in volume to x, 6. weight of water equal in volume to 1 + x. To find the value of x, the following equation, u + + must be solved : - web xxa ( be) =r X c(a - b) !!! Xcla--b) ab - 0) (1 abi mw, y = 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 x = i 1.539 (1.036 — 1.000) 250 X 0.503 125.75 then 2269.6. Answer: 2269.6 Gm. 1.539 X 0.036 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 mw represent the desired weight of the mixture, x 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. (Whenever water is the diluent, c is made 1.000.) Since x + y = mw, and the value of x the liquid to be diluted X cla—b) has been shown above to be the a ( 6-0) latter expression may be substituted for x in the equation, x +y= y X ca-6) +y=mw. This simplified is yca — ycb + yab - yac = mio X a (b — c), and cancelling, y Xb(a —c)=mw X a (b —c). mw X a (b c) ba -c) The value of y (weight of stronger liquid) having been ascertained, it is subtracted from mw, the desired weight of the mixture, to find the value of x, 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 10 X 0.900 (0.960 — 1.000) 0.036 -0.36 0.960 (0.900 — 1.000) 3.75, -0.096 - 0.096 and 10 -3.75 = 6.25. 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 giren 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. 10 x = |