A) 58.25 torr
B) 33 torr
C) 42.1 torr
D) 52.25 torr
Correct Answer: A
Solution :
[a] The given data are \[{{P}_{water}}=17.0\,torr;\] \[{{P}_{total}}\] (4 mole % solution) \[={{P}_{N{{H}_{3}}}}+{{P}_{water}}=50.0\text{ }torr\] \[{{X}_{N{{H}_{3}}}}=0.04\] and \[{{X}_{water}}=0.96\] Now according to Raoult's law; \[{{P}_{water}}={{X}_{water}}P_{water}^{{}^\circ }\] \[=0.96\times 17.0\text{ }torr=16.32\text{ }torr\] Now Henry's law constant for ammonia is \[{{K}_{H}}(N{{H}_{3}})=\frac{{{P}_{N{{H}_{3}}}}}{{{X}_{N{{H}_{3}}}}}=\frac{33.68\,torr}{0.04}=842\,torr\] Hence, for 5 mole % solution, we have \[{{P}_{N{{H}_{3}}}}={{K}_{H}}(N{{H}_{3}}){{X}_{N{{H}_{3}}}}\] \[=\left( 842\,torr \right)\left( 0.05 \right)=16.15\,torr\] Thus, \[{{P}_{total}}\](5 mole % solution) \[={{P}_{N{{H}_{3}}}}+{{P}_{water}}=42.1+16.15=58.25\text{ }torr\]
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