JEE MAIN - Chemistry (2020 - 3rd September Morning Slot - No. 12)
Henry’s constant (in kbar) for four gases $$\alpha $$, $$\beta $$, $$\gamma $$ and $$\delta $$ in water at 298 K is given below :
(density of water = 103 kg m-3 at 298 K)
This table implies that :
| $$\alpha $$ | $$\beta $$ | $$\gamma $$ | $$\delta $$ | |
|---|---|---|---|---|
| KH | 50 | 2 | 2 $$ \times $$ 10-5 | 0.5 |
(density of water = 103 kg m-3 at 298 K)
This table implies that :
solubility of $$\gamma $$ at 308 K is lower than at 298 K
The pressure of a 55.5 molal solution of $$\delta $$ is 250 bar
$$\alpha $$ has the highest solubility in water at a given pressure
The pressure of a 55.5 molal solution of $$\gamma $$ is 1 bar
Explanation
(1) On increasing temperature solubility of gases decreases.
(2) 55.5 molal solution of $$\gamma $$ = 55.5 mol of
P$$\gamma $$ = KHX$$\gamma $$
= 2 $$ \times $$ 10-5 $$ \times $$ $${{55.5} \over {55.5 + {{1000} \over {18}}}}$$
= 10–5 K bar
= 10–2 bar
(3) P$$\delta $$ = KHX$$\delta $$
= 0.5 $$ \times $$ $${{55.5} \over {55.5 + {{1000} \over {18}}}}$$
= 250 bar
(4) With temperature, the value of KH (Henry’s constant) increases and solubility of gas in liquid decreases.
(2) 55.5 molal solution of $$\gamma $$ = 55.5 mol of
P$$\gamma $$ = KHX$$\gamma $$
= 2 $$ \times $$ 10-5 $$ \times $$ $${{55.5} \over {55.5 + {{1000} \over {18}}}}$$
= 10–5 K bar
= 10–2 bar
(3) P$$\delta $$ = KHX$$\delta $$
= 0.5 $$ \times $$ $${{55.5} \over {55.5 + {{1000} \over {18}}}}$$
= 250 bar
(4) With temperature, the value of KH (Henry’s constant) increases and solubility of gas in liquid decreases.
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