Equal masses of ethane (C_2H_6) and hydrogen | vedclass

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(A) Let the mass of ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $w \ g$ each.Molar mass of $C_2H_6 = 30 \ g/mol$.Molar mass of $H_2 = 2 \ g/mol$.Number

Vedclass · Accepted Answer

$15:16$

Vedclass · Answer

(A) Let the mass of ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $w \ g$ each.Molar mass of $C_2H_6 = 30 \ g/mol$.Molar mass of $H_2 = 2 \ g/mol$.Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{w}{30}$.Number of moles of $H_2$ $(n_{H_2})$ = $\frac{w}{2}$.Total moles $(n_{total})$ = $n_{C_2H_6} + n_{H_2} = \frac{w}{30} + \frac{w}{2} = \frac{w + 15w}{30} = \frac{16w}{30}$.Mole fraction of $H_2$ $(X_{H_2})$ = $\frac{n_{H_2}}{n_{total}} = \frac{w/2}{16w/30} = \frac{w}{2} 	imes \frac{30}{16w} = \frac{15}{16}$.According to Dalton's law of partial pressure, the fraction of total pressure exerted by a gas is equal to its mole fraction.Therefore, the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15:16$.

Equal masses of ethane $(C_2H_6)$ and hydrogen $(H_2)$ are mixed in an empty container at $298 \ K$. The fraction of total pressure exerted by hydrogen is:

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