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
B
6×105
C
4×104
D
9×106

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More