A) \[+{{1.86}^{o}}C\]
B) \[-{{1.86}^{o}}C\]
C) \[{{2.42}^{o}}C\]
D) \[{{3.92}^{o}}C\]
Correct Answer: B
Solution :
Key Idea We know that \[\Delta {{T}_{f}}={{K}_{f}}\times \] molality Molarity \[=\frac{weight\text{ }of\text{ }solute\text{ }/\text{ }mol.\text{ }wt.\,\,of\text{ }solute}{weight\,of\,solvent\,(in\,kg)}\] and \[\Delta {{T}_{f}}={{T}_{1}}-{{T}_{2}}\] Here, \[{{T}_{1}}\] is freezing temperature of solvent. Weight of sucrose = 342 g Molecular weight of sucrose \[({{C}_{12}}{{H}_{22}}{{O}_{11}})\] = 342g Weight of solvent = 1000 g = 1 kg Molality \[=\frac{342/342}{1}\] Molality = 1 m \[\because \] \[{{K}_{f}}={{1.86}^{o}}C/mol\] Hence, \[\Delta {{T}_{f}}=1.86\times 1=1.86\] \[\therefore \] \[1.86=0-{{T}_{2}}\] \[{{T}_{2}}=-{{1.86}^{o}}C\] \[\left( \begin{align} & \Delta {{T}_{f}}={{T}_{1}}-{{T}_{2}} \\ & {{T}_{1}}\,of\text{ }water\text{ }is\,{{0}^{o}}C \\ \end{align} \right)\] Therefore, solution will freeze at \[-{{1.86}^{o}}C\].
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