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QJEEM22Ch4338: JEE Main 2022 Chemistry - Class 12 JEEM22Ch4338 Solution: If gas is bubbled through water at , the number of millimoles…

If O2\mathrm{O}_{2} gas is bubbled through water at 303 K303 \mathrm{~K} , the number of millimoles of O2\mathrm{O}_{2} gas that dissolve in 1 litre of water is __________. (Nearest Integer)

(Given : Henry's Law constant for O2\mathrm{O}_{2} at 303 K303 \mathrm{~K} is 46.82 k46.82 \,\mathrm{k} bar and partial pressure of O2=0.920\mathrm{O}_{2}=0.920 bar)

(Assume solubility of O2\mathrm{O}_{2} in water is too small, nearly negligible)

Step-by-Step Verified Solution

P=KH×X\mathrm{P}=\mathrm{K}_{\mathrm{H}} \times \mathrm{X}

0.920bar=46.82×103 bar×mol of O2 mol of H2O0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}

0.920=46.82×103× mol of O21000/180.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}

0.920=46.82×nO20.920=46.82 \times n_{O_{2}}

P=0.92046.82×18=nO2\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}

⇒1.09×10−3 nO2\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}

⇒m mol of O2=1\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1

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