Henry’s constant (in kbar) for four gases $\alpha $, $\beta $, $\gamma $ and $\delta $ in water at 298 K is given below: $\begin{array}{*{20}{c}} {}&\alpha &\beta &\gamma &\delta \\ {{K_H}}&{50}&2&{2 \times {{10}^{ - 5}}}&{0.5} \end{array}$ (density of water $ = {10^3}\,kg\,{m^{ - 3}}$ at 298 K) This implies that

$\alpha $ has the highest solubility in water at a given pressure

Solubility of $\gamma $ at 308 K is lower than at 298 K

The pressure of a 55.5 molal solution of $\gamma $ is 1 bar

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NEET UG CHEMISTRY MCQ: SOLUTIONS

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Henry’s constant (in kbar) for four gases

\alpha

\beta

\gamma

and

\delta

in water at 298 K is given below:

\begin{array}{cccccccccccccccccccc} {}&\alpha &\beta &\gamma &\delta \\ {{K_H}}&{50}&2&{2 \times {{10}^{ - 5}}}&{0.5} \end{array}

(density of water

= {10^3}\,kg\,{m^{ - 3}}

at 298 K)
This implies that

View answer and explanation

Correct option: D

The pressure of a 55.5 molal solution of

\delta

is 250 bar

Option D is correct because according to Henry's law, the partial pressure of a gas is proportional to its mole fraction in the solution. Using the given Henry's constant for δ and the molality, we can calculate the pressure, which is approximately 250 kbar.

NEET UG CHEMISTRY MCQ: SOLUTIONS

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