Search any question & find its solution
Question:
Answered & Verified by Expert
56 g of nitrogen and 96 g of oxygen are mixed isothermally and at a total pressure of 10 atm . The partial pressures of oxygen and nitrogen (in atm) are respectively
Options:
Solution:
1240 Upvotes
Verified Answer
The correct answer is:
6,4
According to Dalton's law,
$p_{\text {solute }}=x_{\text {solute }} \times p$
Given, mass of $\mathrm{N}_2=56 \mathrm{~g}$
$\therefore$ Moles of $\mathrm{N}_2\left(n_{\mathrm{N}_2}\right)=\frac{56}{28}=2 \mathrm{~mol}$
Mass of $\mathrm{O}_2=96 \mathrm{~g}$
$\therefore$ Moles of $\mathrm{O}_2\left(n_{\mathrm{O}_2}\right)=\frac{96}{32}=3 \mathrm{~mol}$
$x_{\mathrm{N}_2}=\frac{2}{2+3}=0.4$
$x_{\mathrm{O}_2}=\frac{3}{2+3}=0.6$
$\therefore \quad p_{\mathrm{N}_2}=x_{\mathrm{N}_2} \times p=0.4 \times 10=4 \mathrm{~atm}$
$p_{\mathrm{O}_2}=x_{\mathrm{O}_2} \times p=0.6 \times 10=6 \mathrm{~atm}$
$p_{\text {solute }}=x_{\text {solute }} \times p$
Given, mass of $\mathrm{N}_2=56 \mathrm{~g}$
$\therefore$ Moles of $\mathrm{N}_2\left(n_{\mathrm{N}_2}\right)=\frac{56}{28}=2 \mathrm{~mol}$
Mass of $\mathrm{O}_2=96 \mathrm{~g}$
$\therefore$ Moles of $\mathrm{O}_2\left(n_{\mathrm{O}_2}\right)=\frac{96}{32}=3 \mathrm{~mol}$
$x_{\mathrm{N}_2}=\frac{2}{2+3}=0.4$
$x_{\mathrm{O}_2}=\frac{3}{2+3}=0.6$
$\therefore \quad p_{\mathrm{N}_2}=x_{\mathrm{N}_2} \times p=0.4 \times 10=4 \mathrm{~atm}$
$p_{\mathrm{O}_2}=x_{\mathrm{O}_2} \times p=0.6 \times 10=6 \mathrm{~atm}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.