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What will be the molarmass of solute at ${27^{\circ} \mathrm{C}}$ solution that contains $3.5 \mathrm{~g}$ of solute in $300.0 \mathrm{ml}$ of solution with an Osmotic pressure of $\mathrm{400pa}$ .
Option: 1
$82.6 \times 10^3$

Option: 2
$7.18 \times 10^4$

Option: 3
$36.41 \times 10^2$

Option: 4
$81.64 \times 10^3$

Answers (1)

best_answer

$T=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$\mathrm{ w_B=3.5 \mathrm{~g} }$
$\mathrm{M_B=?}$
$\mathrm{V_{\text {solution }}=300 \mathrm{ml} }$
$\mathrm{\pi=400 \text { Pascal }=\frac{400}{10^{-5}} \mathrm{~atm} }$

$\mathrm{ \pi=C R T }$
$\mathrm{ \frac{400}{10^5}=\frac{3.5 / M B}{300} \times 1000 \times 0.0821 \times 300 }$
$\mathrm{ M B=7.18 \times 10^4 }$ .

Posted by

Gautam harsolia

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