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Calculate the amount of solute dissolved in 160 gram solvent that boils at $85^{\circ} \mathrm{C}$. the molar mass of solute is $120 \mathrm{~g} \mathrm{~mol}^{-1}$.
$\left(\mathrm{K}_{\mathrm{b}}\right.$ for solvent $=2.7^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$ and boiling point of solvent $\left.=76^{\circ} \mathrm{C}\right)$
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$\left(\mathrm{K}_{\mathrm{b}}\right.$ for solvent $=2.7^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$ and boiling point of solvent $\left.=76^{\circ} \mathrm{C}\right)$
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Verified Answer
The correct answer is:
64 gram
$\Delta \mathrm{T}_{\mathrm{b}}=(85-76)=2.7 \times \frac{\mathrm{W}_{\mathrm{B}}}{120} \times \frac{1000}{160}$
$\mathrm{W}_{\mathrm{B}}=64 \mathrm{~g}$
$\mathrm{W}_{\mathrm{B}}=64 \mathrm{~g}$
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