Solutions Question 331
Question: The total vapour pressure of a 4 mole % solution of $ NH_3 $ in water at 293 K is 50.0 torr. The vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry’s and Raoult’s laws, the total vapour pressure for a 5 mole % solution is
Options:
A) 58.25 torr
B) 33 torr
C) 42.1 torr
D) 52.25 torr
Show Answer
Answer:
Correct Answer: A
Solution:
The given data are $ P_{water}=17.0,torr; $
$ P_{total} $ (4 mole % solution) $ ={P_{NH_3}}+P_{water}=50.0torr $
$ {X_{NH_3}}=0.04 $ and $ X_{water}=0.96 $ Now according to Raoult’s law; $ P_{water}=X_{water}P_water^{{}^\circ } $
$ =0.96\times 17.0torr=16.32torr $ Now Henry’s law constant for ammonia is $ K_{H}(NH_3)=\frac{{P_{NH_3}}}{{X_{NH_3}}}=\frac{33.68,torr}{0.04}=842,torr $ Hence, for 5 mole % solution, we have $ {P_{NH_3}}=K_{H}(NH_3){X_{NH_3}} $
$ =( 842,torr )( 0.05 )=16.15,torr $ Thus, $ P_{total} $ (5 mole % solution) $ ={P_{NH_3}}+P_{water}=42.1+16.15=58.25torr $
