Question

Henry's constant (in kbar) for four gases \(\alpha, \beta, \gamma\) and \(\delta\) in water at 298 K is given below :

	\(\alpha\)	\(\beta\)	\(\gamma\)	\(\delta\)
\(K_H\)	50	2	\(2\times10^{-5}\)	0.5

(density of water = \(10^3\ \mathrm{kg\,m^{-3}}\) at 298 K)
This table implies that :

• A. solubility of \(\gamma\) at 308 K is lower than at 298 K
• B. The pressure of a 55.5 molal solution of \(\delta\) is 250 bar
• C. \(\alpha\) has the highest solubility in water at a given pressure
• D. The pressure of a 55.5 molal solution of \(\gamma\) is 1 bar

Hint:

Use \(p = K_H x\) with \(x_{gas}\) found from 55.5 mol of gas per 1 kg (55.5 mol) of water.

Answer 1

The Correct Option is B

Henry's law: \(p = K_H\,x\). A 55.5 molal solution has 55.5 mol of gas per 1000 g of water, i.e. 55.5 mol of water, so \(x_{gas} = 55.5/(55.5+55.5) = 0.5\).
For \(\delta\): \(K_H = 0.5\) kbar = 500 bar, so \(p = 500\times0.5 = 250\) bar. Option (2) is correct.
For \(\gamma\): \(K_H = 2\times10^{-5}\) kbar = 0.02 bar, \(p = 0.01\) bar, not 1 bar: option (4) is wrong.
Highest solubility at a given pressure means the smallest \(K_H\), which is \(\gamma\), not \(\alpha\): option (3) is wrong.
Option (1) is the general expectation for a gas (solubility falls as temperature rises) but cannot be concluded from this 298 K table; the numerically verifiable choice is (2).
Answer: (2) The pressure of a 55.5 molal solution of \(\delta\) is 250 bar