Calculate the solubility of a gas in water at $260 \ mm \ Hg$ and $25^{\circ} C$ if Henry's law constant of the gas is $0.159 \ mol \ dm^{-3} \ atm^{-1}$ at $25^{\circ} C$.
- A$3.8 \times 10^{-2} \ mol \ dm^{-3}$
- B$5.4 \times 10^{-2} \ mol \ dm^{-3}$
- C$2.7 \times 10^{-2} \ mol \ dm^{-3}$
- D$1.2 \times 10^{-2} \ mol \ dm^{-3}$
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Similar Questions
Observe the following data given in the table $(K_H = \text{Henry's law constant})$. The correct order of solubility of these gases is:
Gas $K_H$ (kbar at $298 \ K$) $CO_2$ $1.67$ $Ar$ $40.3$ $HCHO$ $1.83 \times 10^{-5}$ $CH_4$ $0.413$
| Gas | $K_H$ (kbar at $298 \ K$) |
| $CO_2$ | $1.67$ |
| $Ar$ | $40.3$ |
| $HCHO$ | $1.83 \times 10^{-5}$ |
| $CH_4$ | $0.413$ |
As temperature increases,the value of $K_H$ increases,thus the value of solubility of a gaseous solute will $ . . . . . . $.
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