Question

$250\, g$ solution of $D$-glucose in water contains $10.8 \%$ of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are H, 1u ; C, 12u ; O, 16u)

• A. $1.03$
• B. $2.06$
• C. $3.09$
• D. $5.40$

Answer 1

Answer: (B)

$C _6 H _{12} O _6 \rightarrow \text { Glucose }$
We know: $\frac{\text { mass of } C }{\text { mass of glucose }}=\frac{72}{180}$
Given: $\% C =10.8=\frac{\text { mass of } C }{\text { mass of solution }} \times 100$ $\frac{10.8 \times 250}{100}=$ mass of $C \Rightarrow$ Mass of $C =27$
gm
$\therefore$ mass of glucose $=67.5 gm$
$\therefore$ moles of glucose $=0.375$ moles
Mass of solvent $=250-67.5 gm =182.5 gm$
$\therefore$ Molality $=\frac{0.375}{0.1825}=2.055 \approx 2.06$