What is Henry's law constant of a gas if solubility of gas in water at 25^{\circ} {\C} is 0.028 {\mol} {\dm}^{-3} ? [Partial pressure of gas =0.346 {\bar} ]

Correct Answer is : 0.081moldm−3bar−1

0.081 {\mol} {\dm}^{-3} {\bar}^{-1}

0.075 {\mol} {\dm}^{-3} {\bar}^{-1}

0.093 {\mol} {\dm}^{-3} {\bar}^{-1}

0.049 {\mol} {\dm}^{-3} {\bar}^{-1}

MHTCET 14 May 2023 Shift 2 Solved Paper

Section: Chemistry

Question : 4 of 150

Marks: +1, -0

What is Henry's law constant of a gas if solubility of gas in water at 25∘C is 0.028‌mol‌dm−3 ?
[Partial pressure of gas =0.346‌bar ]

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