Question

$5.5\, mg$ of nitrogen gas dissolves in $180\, g$ of water at $273 \,K$ and one $atm$ pressure due to nitrogen gas. The mole fraction of nitrogen in $180 \,g$ of water at $5 \,atm$ nitrogen pressure is approximately

• A. $1 \times 10^{-5}$
• B. $1 \times 10^{-4}$
• C. $1 \times 10^{-6}$
• D. $1 \times 10^{-3}$

Answer 1

Answer: (B)

Amount of nitrogen gas at 5 atm

$=5 \times 5.5=27.5 mg =27.5 \times 10^{-3}\, g$

Mole fraction of

$N _{2} =\frac{\text { Moles of } N _{2}}{\text { Moles of } H _{2} O +\text { Moles of } N _{2}}$

$=\frac{\text { Moles of } N _{2}}{\text { Moles of } H _{2} O }$

$\left(\because N _{2}< H _{2} O \right)$

$=\frac{27.5 \times 10^{-3}}{\frac{28}{\frac{180}{18}}}=1 \times 10^{-4}$