A) \[0.34{}^\circ C\]
B) \[0.0{}^\circ C\]
C) \[-\,0.34{}^\circ C\]
D) \[-\,0.69{}^\circ C\]
Correct Answer: D
Solution :
Molality of\[{{C}_{2}}{{H}_{5}}OH\](non-electrolyte) solution \[=\frac{17g\text{/}46\times 1000g}{1000\,g\,water}\] \[=0.369\] \[\Delta {{T}_{f}}={{k}_{f}}\times m\] \[=1.86\times 0.369=0.686\] Freezing point of the solution \[=0{}^\circ C-0.686{}^\circ C=-0.69{}^\circ C\]
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