Question

$15 \%$ aqueous solution of glucose (molecular weight = $180\, g / mol$ ) is isotonic with $8 \%$ aqueous solution containing an unknown non-dissociable solute. What is the molecular weight of the unknown solute?

• A. 108
• B. 96
• C. 84
• D. 9.6

Answer 1

Answer: (B)

Given,
Molecular weight of $15 \%$ aqueous solution of glucose $=180\, g / mol$
Molecular weight of solute $=$ ?
$\pi_{1}$ ( glucose ) $=\pi_{2}$ (unknown solute)
$C_{1}$ (urea) $=C_{2}$ (unknown solute)
Now,
${\left[\frac{w_{B} \times 1000}{m_{B} \times V}\right]_{\text {Glucose }}=\left[\frac{w_{B} \times 1000}{m_{B} \times V}\right]_{U}} $
${\left[\frac{15 \times 1000}{180 \times 100}\right]=\left[\frac{8 \times 1000}{m_{B} \times 100}\right]}$
$m_{B}=\frac{180 \times 100 \times 8 \times 1000}{15 \times 1000 \times 100} $
$m_{B}=96$