A mixture of equal masses of ethane (C_2H_6) | vedclass

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(D) Let the mass of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $m \, g$.The molar mass of $C_2H_6 = (2 	imes 12) + (6 	imes 1) = 30 \, g/mol$.Th

Vedclass · Accepted Answer

$15 : 16$

Vedclass · Answer

(D) Let the mass of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $m \, g$.The molar mass of $C_2H_6 = (2 	imes 12) + (6 	imes 1) = 30 \, g/mol$.The molar mass of $H_2 = 2 	imes 1 = 2 \, g/mol$.Number of moles of $C_2H_6$ $(n_1)$ = $m / 30$.Number of moles of $H_2$ $(n_2)$ = $m / 2$.Total moles $(n_{total})$ = $n_1 + n_2 = (m/30) + (m/2) = (m + 15m) / 30 = 16m / 30 = 8m / 15$.The partial pressure fraction of a gas is equal to its mole fraction $(x_i = n_i / n_{total})$.Mole fraction of $H_2$ $(x_{H_2})$ = $(m/2) / (8m/15) = (m/2) 	imes (15/8m) = 15 / 16$.Therefore,the fraction of total pressure exerted by hydrogen is $15/16$.

$A$ mixture of equal masses of ethane $(C_2H_6)$ and hydrogen $(H_2)$ is filled in an empty container at $25\,^oC$. What is the fraction of the total pressure exerted by hydrogen?

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