Question

# Calculate the temperature at which the solution of $$54 \mathrm{g}$$ glucose present in $$250 \mathrm{g}$$ water will be freezed. $$\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$$

Solution

## Given that:Weight of solute $$\left(\mathrm{W}_{\mathrm{B}}\right)=54 \mathrm{g}$$Molecular weight of solute $$\left(\mathrm{M}_{\mathrm{B}}\right)=180 \mathrm{g}$$ molWeight of solvent $$\left(\mathrm{W}_{\mathrm{A}}\right)=250 \mathrm{g}$$$$\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$$Depression in freezing point $$\Delta T_{f}$$$$=\dfrac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}$$$$=\dfrac{1.86 \times 54 \times 1000}{180 \times 250}=2.23 \mathrm{K}$$Depression in freezing point $$=2.23 \mathrm{K}$$Freezing point of solution $$=?$$Freezing point of water $$=273.15 \mathrm{K}$$$$\left(\Delta \mathrm{T}_{\mathrm{f}}\right)=$$ Freezing point of water - Freezing point of solutionFreezing point of solution $$=273.15-2.23=270.92 K$$ChemistryNCERTStandard XII

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