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**Question 1: The density of a 3M sodium thiosulphate solution is 1.25 gm cm ^{–3}. Calculate**

3 moles Na

Wt. of solute Na

wt. of solution = v × d

= 1000 ml × 1.25 gm/ml

= 1000 × 1.25 gm

Wt. of solvent = (1000 × 1.25

Molality = no. of moles of solute per 1000 gm solvent

= 3.865 mol kg

Now, Na

= 7.73 mol kg

Hence molality = 1× 3.865 mol kg

= 3.865 mole/kg

1 L i.e. 1000 ml solution containing 3 moles Na

1000 × 1.25 gm solution containing 3 ×158 gm Na

Wt. of Glauber’s satt = w

= 80.575 gm

density of solution = d = 1077.2 kg m

P

The composition of the vapour in mole fraction units can be obtained by using Dalton’s law. In vapour phase we have;

Again

Now the mole fraction of benzene and toluene in liquid are 0.63 and 0.37 respectively. Therefore, their vapour pressures are given by

P

Now we can calculate the mole fractions in vapour phase to be

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