Solutions Question 259

Question: Freezing point of a biological fluid is $ -0.60{}^\circ C $ in aqueous solution $ K_{f}(H_2O)=1.86{}^\circ mo{l^{-}}~kg. $ Thus, its osmotic pressure at 310 K is (assume molarity = molality)

Options:

A) 0.0766 atm

B) 7.66 atm

C) 0.766 atm

D) 8.19 atm

Show Answer

Answer:

Correct Answer: D

Solution:

$ \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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