Question

The boiling point of a solution made by dissolving 12.0 g of glucose in 100 g of water is ${100.34}^{\circ }C$. Calculate the molecular weight of glucose; $K_b$ for water $={0.52}^{\circ }C/m$.

• A. $183.5gmo{l}^{-}1$
• B. $283.5gmo{l}^{-}1$
• C. $383.5gmo{l}^{-}1$
• D. $73.5gmo{l}^{-}1$

Answer 1

The correct option is A $183.5gmo{l}^{-}1$

Using the relation for the molecular weight of a solute from elevation in boiling point, we have
$M{w}_B=K_b(\frac{W_B}{W_A\Delta T_b}\times 1000)0.52(\frac{12}{100\times 0.34}\times 1000)(\Delta T_b=100.34-100={0.34}^{\circ }C)\Rightarrow M{W}_B=183.5gmo{l}^{-1}$