Test Index

NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions

© examsnet.com
Question : 35 of 53
Marks: +1, -0
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27×1054.27 \times 10^{5} mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Solution:  
Here, KH=4.27×105K_H = 4.27 \times 10^{5} mm Hg, p = 760 mm Hg
Applying Henry’s law, p=KHxp=K_H x
∴x=pKH=7604.27×105\therefore x=\frac{p}{K_H}=\frac{760}{4.27 \times 10^{5}} =1.78×10−3=1.78 \times 10^{-3}
i.e., solubility in terms of mole fraction of methane in benzene =1.78×10−3= 1.78 \times 10^{-3}
© examsnet.com
Go to Question:
Prev Question
Next Question