A) \[\text{ }7.06{}^\circ C\]
B) \[3.51{}^\circ C~\]
C) \[\text{ }3.51{}^\circ C\]
D) \[10{}^\circ C\]
Correct Answer: A
Solution :
10% solution means, 1. 00 kg water contains 0.100 kg NaCl and 0.900 kg \[{{H}_{2}}O\]) \[\therefore \] \[(100\,g\,NaCl)\times \frac{1\,mol}{58.5\,g}=1.709\,mol\] \[=\frac{1.709\,mol}{0.900\,kg}=1.89\,m\] For NaCl, i = 2 \[(\because \,\,NaCl\,\,\to \,N{{a}^{+}}+C{{l}^{-}})\] \[\Delta {{T}_{f}}=i\,Kf\cdot m\] \[=2\times ({{1.86}^{o}}C/m)\times (1.89\,m)\] Freeezing point of the solution, \[{{T}_{b}}={{T}^{o}}-\Delta {{T}_{f}}\] \[={{0}^{o}}-{{7.06}^{o}}C=-{{7.06}^{o}}C\]
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