The total pressure of a mixture of non-reacti | vedclass

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(D) Step $1$: Calculate the number of moles of each gas.$n_X = \frac{0.6 \ g}{20 \ g \ mol^{-1}} = 0.03 \ mol$$n_Y = \frac{0.45 \ g}{45 \ g \ mol^{-1}

Vedclass · Accepted Answer

$555$

Vedclass · Answer

(D) Step $1$: Calculate the number of moles of each gas.$n_X = \frac{0.6 \ g}{20 \ g \ mol^{-1}} = 0.03 \ mol$$n_Y = \frac{0.45 \ g}{45 \ g \ mol^{-1}} = 0.01 \ mol$Step $2$: Calculate the mole fraction of gas $X$ $(\chi_X)$.$\chi_X = \frac{n_X}{n_X + n_Y} = \frac{0.03}{0.03 + 0.01} = \frac{0.03}{0.04} = 0.75$Step $3$: Calculate the partial pressure of gas $X$ $(P_X)$.$P_X = \chi_X 	imes P_{total} = 0.75 	imes 740 \ mm \ Hg = 555 \ mm \ Hg$

The total pressure of a mixture of non-reacting gases $X$ $(0.6 \ g)$ and $Y$ $(0.45 \ g)$ in a vessel is $740 \ mm$ of $Hg$. The partial pressure of the gas $X$ is $........ \ mm$ of $Hg$. (Nearest Integer) (Given: molar mass $X = 20 \ g \ mol^{-1}$ and $Y = 45 \ g \ mol^{-1}$)

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