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$15 \%$ aqueous solution of glucose (molecular weight $=180 \mathrm{~g} / \mathrm{mol}$ ) is isotonic with $8 \%$ aqueous solution containing an unknown non-dissociable solute. What is the molecular weight of the unknown solute?
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The correct answer is:
96
Given,
Molecular weight of $15 \%$ aqueous solution of glucose $=180 \mathrm{~g} / \mathrm{mol}$
Molecular weight of solute $=$ ?
$\pi_1($ glucose $)=\pi_2($ unknown solute $)$
$C_1($ urea $)=C_2($ unknown solute $)$
Now, $\left[\frac{w_B \times 1000}{m_B \times V}\right]_{\text {Glucose }}=\left[\frac{w_B \times 1000}{m_B \times V}\right]_{\text {Unknown solute }}$
$\left[\frac{15 \times 1000}{180 \times 100}\right]=\left[\frac{8 \times 1000}{m_B \times 100}\right]$
$m_B=\frac{180 \times 100 \times 8 \times 1000}{15 \times 1000 \times 100}$
$m_B=96$
Molecular weight of $15 \%$ aqueous solution of glucose $=180 \mathrm{~g} / \mathrm{mol}$
Molecular weight of solute $=$ ?
$\pi_1($ glucose $)=\pi_2($ unknown solute $)$
$C_1($ urea $)=C_2($ unknown solute $)$
Now, $\left[\frac{w_B \times 1000}{m_B \times V}\right]_{\text {Glucose }}=\left[\frac{w_B \times 1000}{m_B \times V}\right]_{\text {Unknown solute }}$
$\left[\frac{15 \times 1000}{180 \times 100}\right]=\left[\frac{8 \times 1000}{m_B \times 100}\right]$
$m_B=\frac{180 \times 100 \times 8 \times 1000}{15 \times 1000 \times 100}$
$m_B=96$
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