Solvent X has an unknown molal elevation constant. Dissolving $1.8 g$ of glucose (M $= 180 g mo l^{- 1}$ ) in $50 g$ of X raises its boiling point by $0.40 K$ . A second solution, $3.0 g$ of an unknown non-electrolyte in $60 g$ of X, raises the boiling point by $0.25 K$ . The molar mass of the unknown solute (in $g mo l^{- 1}$ ) is

Answer: B

From glucose:

m = (1.8/180) /0.050 = 0.200 mol k g^{- 1}

, so

K_{b} = 0.40/0.200 = 2.00 K kg mo l^{- 1}

. For the unknown,

m = 0.25/2.00 = 0.125 mol k g^{- 1}

, so moles

= 0.125 \times 0.060 = 0.0075

. Then

M = 3.0/0.0075 = 400 g mo l^{- 1}