A) 0.0766 atm
B) 7.66 atm
C) 0.766 atm
D) 8.19 atm
Correct Answer: D
Solution :
[d] \[\Delta {{T}_{f}}=molality\times {{K}_{f}}\] \[0.60{}^\circ =molality\times 1.86{}^\circ mo{{l}^{-1}}kg\]. Given molality = molarity \[\therefore Molality=\frac{0.60}{1.86}=0.322mol\,k{{g}^{-1}}\] Or molarity\[=0.301\,mol\,{{L}^{-1}}\] Also, \[\pi =MRT=0.322\times 0.0821\times 310=8.19atm\]
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