A) \[4.16\times {{10}^{-3}}\]
B) \[2.38\times {{10}^{-6}}\]
C) \[6.43\times {{10}^{-3}}\]
D) \[3.78\times {{10}^{-6}}\]
Correct Answer: A
Solution :
From neutralization reaction, meq of acid \[=125\times 0.1=12.5\] Eq. wt. of acid \[=\frac{1000}{12.5}=80\] mol. wt. Molarity \[=\frac{1}{80}\times \frac{100}{15}=0.083\] \[-\Delta {{T}_{f}}=i{{k}_{f}}m=(1+\alpha )\times 1.86\times 0.083\] \[\alpha =0.2\] \[{{k}_{a}}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{0.093\times {{(0.2)}^{2}}}{0.8}=4.16\times {{10}^{-3}}\]
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