Question

$250g$ 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)

$C_6{H}_{12}{O}_6\to \text{ Glucose }$
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 $=67.5gm$
$\therefore$ moles of glucose $=0.375$ moles
Mass of solvent $=250-67.5gm=182.5gm$
$\therefore$ Molality $=\frac{0.375}{0.1825}=2.055\approx 2.06$