A) 2.4 g
B) 1.8 g
C) 1.0 g
D) 1.5 g
Correct Answer: A
Solution :
\[\Delta {{T}_{f}}=i{{k}_{f}}.m\] ?(i) where, m = molality \[\Delta {{T}_{f}}=\]depression in freezing point i = van?t Hoff factor and \[m=\frac{w}{122}\times \frac{1000}{30}\] For association,\[i=1+\left( \frac{1}{2}-1 \right)0.8=0.6\] So, form eqn. (i), \[2=0.6\times 5\times \frac{w}{122}\times \frac{1000}{30}\] or\[w=\frac{122\times 2\times 30}{0.6\times 5\times 100}=2.44g\]You need to login to perform this action.
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