A non-volatile solute ( $12.0$ g) dissolved in $90.0$ g of water ( $18 g mo l^{- 1}$ ) gives a relative lowering of vapour pressure of $0.0500$ . Do not use the dilute approximation. The molar mass of the solute (in $g mo l^{- 1}$ ) is closest to

Answer: B

Relative lowering

= x_{2} = 0.0500

, so

n_{2} / (n_{2} + 5.00) = 0.0500

. This gives

n_{2} = 0.0500 \times 5.00/0.950 = 0.2632

mol.

M = 12.0/0.2632 = 45.6 g mo l^{- 1}

. The dilute form

n_{2} / n_{1}

would give 48.0, which is not exact here.