Question

$15\%$ aqueous solution of glucose (molecular weight = $180g/mol$ ) 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 $=180g/mol$
Molecular weight of solute $=$ ?
${\pi }_1$ ( glucose ) $={\pi }_2$ (unknown solute)
$C_1$ (urea) $=C_2$ (unknown solute)
Now,
${\left[\frac{w_B\times 1000}{m_B\times V}\right]}_{\text{Glucose }}={\left[\frac{w_B\times 1000}{m_B\times V}\right]}_U$
$\left[\frac{15\times 1000}{180\times 100}\right]=\left[\frac{8\times 1000}{m_B\times 100}\right]$
$m_B=\frac{180\times 100\times 8\times 1000}{15\times 1000\times 100}$
$m_B=96$