A) 28.5%
B) 32.6%
C) 73.3%
D) 82%
Correct Answer: C
Solution :
[c] Observed mol. mass\[=\frac{100\times {{k}_{f}}\times w}{W\times \Delta {{T}_{f}}}\] \[=\frac{1000\times 5.12\times 20\times {{10}^{-3}}}{1\times 0.69}=148.4\] Normal mol, mass of phenol=94 \[i=\frac{\text{Normal}\,\,\text{mol}\text{.}\,\,\text{mass}}{\text{Observed}\,\,\text{mol}\text{.}\,\,\text{mass}}=\frac{94}{148.4}=0.633\]Since, phenol is dimerized \[\therefore n=2\] Also, \[i=1+\left[ \frac{1}{n}-1 \right]\alpha =1+\left[ \frac{1}{2}-1 \right]\alpha =1-\frac{\alpha }{2}\] \[\Rightarrow 0.633=1-\frac{\alpha }{2}\] \[\Rightarrow \frac{\alpha }{2}=0.367\] \[\Rightarrow \alpha =0.733\times 100=73.3%\]
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