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 nitorgen (in atm) are respectively

Options

A. 4, 6
B. 5, 5
C. 2, 8
D. 6, 4

Answer

D

Explanation

According to Dalton.s law,
`p_("solute") = x_("solute") xx p`
where, `p_("solute")` = partial pressure of solute
`x_("solute")` = mole fraction of solute
p = total pressure
Given, mass of `N_(2) = 56 g`
`therefore` Moles of `N_(2)(""^(n)N_(2)) = (56)/(28) = 2` mol
Mass of `O_(2) = 96 g`
`therefore` Moles of `O_(2)(""^(n)O_(2)) = (96)/(32) = 3` mol
`x_(N_(2)) = (2)/(2+3) = 0.4`
`x_(O_(2)) = (3)/(2+3) = 0.6`
`therefore p_(N_(2)) = x_(N_(2)) xx p = 0.4 xx 10 = 4` atm
`p_(O_(2)) = x_(O_(2)) xx p = 0.6 xx 10 = 6` atm