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Difference of level, 140 ̊5 mm.

Coefficient of expan

sion of mercury

=

0'0001815.

Coefficient of cubical expansion of glass, o'0000262.

255. 15055 grams of a mixture of sodium and potassium chlorides gave 3'4222 grams of silver chloride. Calculate the relative amounts of the two chlorides.

256. 12060 grams of a mixture of sodium bromide and sodium chloride gave, on complete precipitation by silver nitrate, 2.6554 grams of mixed silver chloride and bromide, which, on reduction, yielded 18418 grams of metallic silver. Calculate the proportion of chloride and bromide.

SOLUTION OF GASES IN LIQUIDS.

GASES fall into two groups when considered in relation to their solubility in liquids :

(1). Those gases completely expelled from the solvent by raising its temperature or reducing the pressure to which it is subjected.

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I. In group (1), the amount of gas absorbed depends upon :

I. The nature of the gas and of the absorbing liquid.

II. The actual pressure of the gas considered. Henry's Law :-The quantity of any gas absorbed by a given quantity of a liquid is proportional to the pressure.

III. The temperature. Volume absorbed decreases with the increase of temperature; it may be expressed by means of an empirical formula of the form V-a-bt+ct2, where a, b, c are constants for each gas and is the temperature.

The absorption-coefficient of a gas is a number which expresses the ratio of the volume of the gas, measured at 0° C. and the pressure of absorption, absorbed by the liquid at the temperature of observation to the volume of

the absorbing liquid. In a mixture of gases, each component behaves as if it alone were present under the partial pressure which is exerted by itself.

EXAMPLE.

It is required to calculate the absorption coefficient of nitrogen dissolved in water at 19° C. from the appended data (Bunsen).

I. Observations before Absorption.

Lower level of mercury in outer cylinder_a =
Upper level of mercury in absorption

tube

423.6 mm.

Height of barometer

Temperature of absorptiometer

Temperature of barometer

=

=

p

t

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II. Observations after Absorption.

Lower level of mercury in outer cylinder Upper level of mercury in absorption tube

Water level in absorption tube.

Water level in outer cylinder
Height of barometer

Temperature of absorptiometer

Temperature of barometer

III. Tabular Data.

a1

=

352*2 mm.

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Volumes, according to the calibration table, of:

Gas taken. mark b = 124'1 mm.

Tube above b1

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Tube above 1 = 65.5 mm., remaining gas.
Hence water used for absorption.

Coefficient of expansion of mercury for 1° C..

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=

34'90. 200'04. 17.67. €182.37.

=0'00018.

P and P1 being the pressures of the dry gas before and after absorption respectively, V and V1 being the corresponding volumes of gas reduced to oo C.

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257. The absorption coefficient of nitrogen dissolved in water is oo152 at 12.6° C.; what volume of the gas measured at o° C. and 760 mm. pressure is absorbed by one litre of water at 12.6° C. at each of the pressures :

1000 mm., 748°2 mm., 391 mm., and 14'3 mm. ?

258. What is the volume of nitrogen absorbed in each case in question 257, measured at the temperature and pressure of the experiment? What is the weight of the gas absorbed in each case?

259. The absorption coefficient of hydrogen dissolved in water is represented by the interpolation formula 0*0215286 000019216 +00000017228t2: what quantities of the gas measured (a) under the experimental conditions, (b) under standard conditions,

a =

will be given off on boiling 325 c.c. of the solution made by agitating water in hydrogen (A) at 10° C. and 750 mm., (B) at 14° C. and 767 mm., (C) at 18° C. and 732 mm. ?

260. The coefficient of absorption of carbon dioxide in water is found to be a = 17967 - 007761 +0001642412; find the coefficients for t 44° C., 84° C., 13.8° C., 16·6° C., 19′1° C., and 22′4° C.

=

261. The coefficient of absorption of hydrogen in alcohol is given by the interpolation formula a=0'069250'0001487 +00000012; find the values of the coefficient at 1°, 5°, 11'4°, 14'4°, and 19′9° C., and compare each value with the corresponding value of the coefficient for water as given in question 259.

262. Find an interpolation formula expressing the variation of the coefficient of absorption of carbonic oxide in water with alteration of temperature, the coefficients found by experiment being :—at 5·8°, 0028636; at 8.6°, 0027125; at 174°, 0023854; at 184°, o'023147; at 9°, 0026855; and at 22°, 0'022907.

263. At 23° C. the absorption-coefficient of oxygen in water is 0'03402, that of nitrogen is given by the expression a = 00203460000538871 +00000111562; what will be the percentage composition by volume of the gas mixture obtained by boiling water previously saturated with air at 23° C. ? (Air :-N, 79′1; O, 20 ̊9.)

264. A litre of water saturated with carbon dioxide at 4‍4° C. and 748 mm. is shaken up with a litre of nitrogen at 23° C. and 760 mm.; the temperature of the whole being now 23° C., what is the composition of the gas remaining over the solution, a for nitrogen being taken as in question 263, and for CO2 being equal to 17967 – 0'077611+0'0016424/2?

265. A mixture of hydrogen and carbon dioxide is agitated with 356 ̊4 c.c. of water at 5·5° C., volume of gas before absorption reduced to o° C. 171 29 c.c. and pressure = o'5368 m.; vol. of gas after absorption at o° C 11961 c.c., and its pressure

=

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o'6809 m.; a for H 00193, for CO2 14199; what is the percentage composition of the original mixture?

=

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SOLUBILITY OF SOLIDS IN LIQUIDS.
MOLECULAR WEIGHT AND

THE

LOWERING OF THE FREEZING-POINT
OF SOLUTIONS.

THE solubility of solids in liquids does not admit of being represented by any simple law; the alteration of the solubility with temperature may generally be expressed by an equation of the form = a + bt + ct2 + dt3, where a, b, c, d are constants for each set of substances considered, and t is the temperature.

As these constants have only been determined in a few cases, the consideration of problems arising in connection therewith does not fall within the scope of this book.

It has been found that substances dissolved in any solvents which solidify at attainable temperatures cause a lowering of the freezing-point according to the following law :

If the molecular weight (in terms of any unit) of any substance be dissolved in 100 times the molecular weight (in terms of the same unit) of any liquid, the freezingpoint of the latter is lowered by an amount always very near to o'63° C.

Let A be the coefficient of lowering of the temperature of solidification (lowering produced by one gram dissolved in 100 grams of solvent), M be the molecular weight of the dissolved compound, T be the molecular lowering of the freezing-point (that produced by the molecular weight in grams of the substance dissolved in 100 grams of the T solvent); then MA=T, and M

=

Α

If P be the weight of the solvent, P' the weight of the dissolved body, K the lowering of the freezing-point given by experiment, we have

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