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$12 \mathrm{~g}$ of a nonvolatile solute dissolved in $108 \mathrm{~g}$ of water produces the relative lowering of vapour pressure of $0.1$. The molecular mass of the solute is :
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$20$
$20$
$\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}_s}{\mathrm{P}^{\mathrm{o}}}=\frac{n}{N}=\frac{w}{m} \times \frac{\mathrm{M}}{W}$
$0.1=\frac{12}{m} \times \frac{18}{108}$
$m=\frac{12 \times 18}{0.1 \times 108}=20$
$0.1=\frac{12}{m} \times \frac{18}{108}$
$m=\frac{12 \times 18}{0.1 \times 108}=20$
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