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What is te mass of solute having molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ when dissolved in 98 gram solvent decreases its freezing point by $0.2 \mathrm{~K}$ ?
(The numerical value of cryoscopic constant of solvent is 1.71)
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(The numerical value of cryoscopic constant of solvent is 1.71)
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Verified Answer
The correct answer is:
0.687 gram
$\Delta \mathrm{Tf}=\mathrm{K}_{\mathrm{f}} \mathrm{m}$
$0.2=1.71 \times \frac{\mathrm{W}_{\mathrm{B}}}{60} \times \frac{1000}{98}$
$\mathrm{W}_{\mathrm{B}}=0.687 \mathrm{~g}$
$0.2=1.71 \times \frac{\mathrm{W}_{\mathrm{B}}}{60} \times \frac{1000}{98}$
$\mathrm{W}_{\mathrm{B}}=0.687 \mathrm{~g}$
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