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Grade 12Physical Chemistry

How many grams of NaBr must be added to 270g of water to lower the vapour pressure by 3.125mmHg.at a temperature at which vapour pressure of water is 50mmHg (at. Mass of Na =23 , Br= 80).Assume 100% ionisation of NaBr

a.151.5g. b.0.51g. C. 5.5g. D.1.5g. E.51.5g

Bhavya

7 Years agoGrade 12

1 Answer

Samyak Jain

7 Years ago

According Roult’s law, for non-electrolyte solutions

$\Delta$ P / P₀ = X_B, where X_B is the mole fraction of solute in solution = n_B / (n_A + n_B)

A represents solvent and B represents solute.

For dilute solutions, X_B $\approx$ n_B / n_A .

For dilute electrolyte solutions, the formula becomes

$\Delta$ P / P₀ = i.n_B / n_A , where i is Vant Hoff factor.

Here i = 2, P₀ = 50 mm Hg, $\Delta$ P = 3.125 mm Hg, n_A = 270 g / (18 g/mol) = 16 mol.

$\therefore$ 3.125 / 50 = 2.n_B / 16 or 0.0625 = n_B / 8

So, n_B = 0.5 = Required mass of NaBr / Molar mass of NaBr

Thus, Required mass of NaBr = 0.5 x 103 = 51.5 g

Option E is correct.

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