A) \[80%\]
B) \[60%\]
C) \[65%\]
D) \[75%\]
Correct Answer: A
Solution :
: Given: \[{{K}_{f}}=1.86K\text{ }kg\text{ }mo{{l}^{-1}},w=1.25\text{ }g,\] \[~W=50g,Ar={{0.3}^{o}}C,M=?\] As P undergoes association, \[2P{{(P)}_{2}};n=2\] Using equation, \[M=\frac{1000\times {{K}_{f}}\times w}{W\times \Delta T}\] \[M=\frac{1000\times 1.86\times 1.25}{50\times 0.3}=155\] Now, \[i=\frac{Normal\,mol.\,mass}{Observed\,mol.\,mass}=\frac{94}{155}=0.606\] Then, the degree of association of P is \[\alpha =\frac{1-i}{1-\frac{1}{n}}\] \[\alpha =\frac{1-0.606}{1-\frac{1}{2}}=0.788\] or \[78.8%\sim 80%\]You need to login to perform this action.
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