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Question

The Henry's law constant for the solubility of $${N}_{2}$$ gas in water at 298 K is $$1.0 \times {10}^{5}\ atm$$. The mole fraction of $${N}_{2}$$ in air is 0.8. The no. of moles of $${N}_{2}$$ from air dissolved in 10 moles  of water of 298 K and 5 atm pressure is:


A
2×105
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B
6×105
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C
4×104
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D
9×106
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Solution

The correct option is C $$4 \times {10}^{-4}$$
$$ \begin{array}{l} \text { Alc to Henry's law, } P=K x \\ \qquad \begin{array}{l} x \text { is the mole fraction } \\ k \text { is the Henry's law constant } \\ P \text { is partial pressure in atm } \end{array} \end{array} $$
$$\begin{array}{l}\text { Substituing the values, }\\\qquad \begin{array}{l}c=\frac{4}{1 \times 10^{5}}=4 \times 10^{-5} \\\text {no. of moles of } N_{2} \text { in } 10 \text { moles of water }\\\frac{n}{10}=4 \times 10^{-5}\\\Rightarrow n=4\times 10^{-4}\mathrm{~mol}\end{array}\end{array}$$
Hence ,(C) is correct option.

Chemistry

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