$2.0 g$ of a non-electrolyte X dissolved in $80 g$ of water freezes at $- 0.465^{\circ} C$ . Using $K_{f} = 1.86$ and $K_{b} = 0.52 K kg mo l^{- 1}$ (normal boiling point of water $100.00^{\circ} C$ ), the boiling point of a solution of $5.0 g$ of X in $100 g$ of water is

Answer: A

From freezing:

m = 0.465/1.86 = 0.250 mol k g^{- 1}

, moles of X

= 0.250 \times 0.080 = 0.020

, so

M = 2.0/0.020 = 100 g mo l^{- 1}

. For the second solution

m = (5.0/100) /0.100 = 0.50 mol k g^{- 1}

and

Δ T_{b} = 0.52 \times 0.50 = 0.26 K

, so the boiling point is

100.26^{\circ} C