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Chemistry Class 12 NCERT Solutions: Chapter 2 Solutions Part 15
Q: 35. Henry՚s law constant for the molality of methane in benzene at mm Hg. Calculate the solubility of methane in benzene at Kunder .
Answer:
Here,
According to Henry՚s law,
Hence, the mole fraction of methane in benzene is .
Q: 36.100 g of liquid A was dissolved in of liquid B (molar mass 180 g mol-1) . The vapour pressure of pure liquid B was found to be 500 Torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:
Number of moles of liquid A,
Number of moles of liquid B,
Then, mole fraction of A,
And, mole fraction of B,
Vapour pressure of pure liquid B, torr
Therefore, vapour pressure of liquid B in the solution,
Total vapour pressure of the solution,
Vapour pressure of liquid A on the solution,
Now,
Hence, the vapour pressure of pure liquid A is 280.7 torr
Q: 37. Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot and a function of . The experimental data observed for different compositions of mixture is.
0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 | |
0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 | |
632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
Answer:
From the question, we have the following data
0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 | |
0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 | |
632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 | |
632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8 |
It can be observed from the graph that the plot for the of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.