Solutions Question 160
Question: At $ 20{}^\circ C $ and 1.00 atm partial pressure of $ H_2 $ , 18 mL of $ H_2 $ (STP) dissolves in 1 L of water. If 2 L of water is exposed to a gaseous mixture having a total pressure of 1425 ton- (excluding the vapour pressure of water) and containing 80% $ H_2 $ by volume, the volume of $ H_2 $ (STP) dissolved is
Options:
A) 27 mL
B) 54 mL
C) 33.75 mL
D) 67.50 mL
Show Answer
Answer:
Correct Answer: B
Solution:
$ V_{dissolved}=KP_{H2} $
$ P_{H2} $ in the mixture $ =X_{H2}P=\frac{80\times 1425}{100}=1140 $
$ torr=\frac{1140}{760}=1.5atm $ Hence, $ V_{dissolved} $ in 2L of water $ =2\times K\times P_{H2} $
$ =2\times 18\times 1.5=54mL $
BETA
