Question

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34°C.

Answer 1

$\boxed{ \triangle{} T _{f} = K _{f} \: . \: m} \\ \\ \triangle{} T _{f} = \: ( \: Original \: freezing \: point \: of \: \\ \: \: \: \: \: \: water \: ) \: - \: ( \: freezing \: point \: of \: solution \: ) \\ \\ \triangle{} T _{f} = \: 0.340 \: K \\ \\ m \: = \frac{given \: mass \: of \: solute}{molar \: mass \: of \: solute} \times \frac{1000}{mass \: of \: solvent} \\ \\ m = \: 33.3 \: g \: Kg {}^{ - 1} M {}^{ - 1} \\ \\ K _{f} \: \: of \: water \: = \: 1.86 \: K \: \: Kg {}^{ - 1} mol {}^{ - 1} \\ \\ Putting \: values \: , \: we \: get \: - \\ \\ \color{purple} \large \boxed{ M \: = \: 182.17 \: g \: mol {}^{ - 1} }$

Answer 2

Explanation:

molecular mass is 18 2.35 gram per mole