JEE MAIN - Chemistry (2021 - 25th July Morning Shift - No. 18)
CO2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2 exerts a partial pressure of 0.835 bar then x m mol of CO2 would dissolve in 0.9 L of water. The value of x is ____________. (Nearest integer)
(Henry's law constant for CO2 at 298 K is 1.67 $$\times$$ 103 bar)
(Henry's law constant for CO2 at 298 K is 1.67 $$\times$$ 103 bar)
Answer
25
Explanation
From Henry's law
Pgas = KH.Xgas
0.835 = 1.67 $$\times$$ 103 $$\times$$ $${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$$
$$ \Rightarrow $$ n(CO2) = 0.025
Millimoles of CO2 = 0.025 $$\times$$ 1000 = 25
Pgas = KH.Xgas
0.835 = 1.67 $$\times$$ 103 $$\times$$ $${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$$
$$ \Rightarrow $$ n(CO2) = 0.025
Millimoles of CO2 = 0.025 $$\times$$ 1000 = 25
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