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Chemistry >> Physical Chemistry >> Solutions IIT JEE

What approximate proportions by volume of water( d=1g/cc) and ethylene glycol C2H6O2 (d=1.12g/cc) must be mixed to ensure protection of an automobile cooling system to -10 C?

Asked By: AMIT DAS
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Joshi sir comment

we know that  dT = Kfm so m = 10/1.86

m means moles of ethelene glycol in 1000 gm water

so 1000 gm water contains (10/1.86) * 62 gm ethylene glycol

so 1000 cc water contains [(10/1.86)* 62] / 1.12 cc

now calculate

Chemistry >> Physical Chemistry >> Solutions IIT JEE

10.1 g of a volatile liquid occupies a volume of 4 litres when vaporised at 100 C and 70 cm pressure. what would be the osmotic pressure of a 2% ( grams per 100 cc) solution of this substance  at 0 C?

Asked By: AMIT DAS
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Joshi sir comment

10.1 g of a volatile liquid occupies a volume of 4 litres when vaporised at 100 C and 70 cm pressure.

first convert this volume to STP by using the formula P1V1/T1 = P2V2/T2

after getting the value of  V2 , we can say that V2 volume contains 10.1 gm so 22.4 litre will contain 10.1*22.4/V2

thus we will get molar mass

now use πV = (w/M) RT

V = 0.1 litre, w = 2 gms, M = calculated above, T = 273 K

Chemistry >> Physical Chemistry >> Solutions IIT JEE

The vapour pressure of pure benzene is 22 mm & that of pure toulene is 75 mm at 200  C . what is the composition of the solution of these two components that has a vapour pressure of 50 mm at this temperaure? what is the composition of vapour in equilibrium with this solution?

Asked By: AMIT DAS
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Joshi sir comment

x22+(1-x)75 = 50 

so 22x+75-75x = 50

so 53x = 25 so x = 25/53

so ratio of there amounts (in mole) = 25/53 : 28//53    =  25:28

now ratio in vapour form can be obtained by this method

v. p. of pure benzene = 25/53 * 22 = 550/53

and v. p. of pure toluene = 28/53 * 75 = 2100/53

so ratio = (550/53)/(2100/53) = 550/2100 = 55/210 = 11/42

Chemistry >> Physical Chemistry >> Solutions IIT JEE
depression in freezing point of 0.01 molal acetic acid solution is 0.0204 deg.cel. 1molal of this solution freezes at -1.86deg.cel.assuming molality equals molarity find pH of acid A. 2 B. 3 C. 3.2 D. 4.2
Asked By: GAURAV MAHATE
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Joshi sir comment

for 1 molal soln. depression = 2.04 deg. cel.

and normal depression = 1.86

so vant hoff factor = 2.04/1.86 = 102/93 = 34/31

now CH3COOH --------->   CH3COO-  +  H+

let degree of dissociation = α                 

so concentrations after dissociation will be 

       1-α                                   α                  α

so according to these concentrations vant hoff factor = 1+α/1

on comparing 1+α = 34/31 or α = 3/31               so pH = -log3/31

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