Moles of K2SO4 to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is

Question

Moles of K2SO4 to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is (A) 3 mol (B) 2 mol (C) 1 mol (D) 0.5 mol

Tardigrade Admin · Accepted Answer

Lowering ofV.P.is colligative property thus, iK2 SO 4=1+(y-1) x=1+2 x=3 ∴ text If (∅p/P*)=(n1 i/n1 i+n2) (10/50)=(3 n1/3 n1+12)=(n1/n1+4) n1=1

Q. Moles of $K_{2} S O_{4}$ to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is

3 mol

2 mol

1 mol

0.5 mol

Lowering ofV.P.is colligative property thus, $i_{K_{2}} S O_{4} = 1 + (y - 1) x = 1 + 2 x = 3$
$∴ If \frac{\emptyset _{p}}{P ^{*}} = \frac{n _{1} i}{n _{1} i + n _{2}}$
$\frac{10}{50} = \frac{3 n _{1}}{3 n _{1} + 12} = \frac{n _{1}}{n _{1} + 4} n_{1} = 1$

Q. Moles of K2​SO4​ to be dissolved in 12 moles of water to lower its vapour pressure by 10 mm Hgat a temperature at which vapour pressure of pure water is 50 mmHg is