Solutions Question 88
Question: Freezing point of a biological fluid is $ -0.60{}^\circ C $ in aqueous solution $ K_{f}(H_2O)=1.86{}^\circ mo{l^{-1}}~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 $
BETA
