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(D) Let the weight of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $W \ g$.Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{W}{30}$.Number of mol

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$15: 16$

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(D) Let the weight of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $W \ g$.Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{W}{30}$.Number of moles of $H_2$ $(n_{H_2})$ = $\frac{W}{2}$.According to Dalton's law of partial pressures, the fraction of total pressure exerted by a gas is equal to its mole fraction $(\chi)$.$\chi_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{C_2H_6}} = \frac{\frac{W}{2}}{\frac{W}{2} + \frac{W}{30}}$.$\chi_{H_2} = \frac{\frac{W}{2}}{\frac{15W + W}{30}} = \frac{W}{2} 	imes \frac{30}{16W} = \frac{15}{16}$.Thus, the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15: 16$.

Equal weights of ethane and hydrogen are mixed in an empty container at $25^{\circ} C$. The fraction of total pressure exerted by hydrogen is

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