Find the fraction of the total pressure exerted by hydrogen if it is mixed with ethane in an empty container at 25℃

The correct option is 15/16.

To understand why this is the correct fraction, let us analyze the properties of the gases involved in the mixture under the given conditions.
When two gases are mixed in a container, their pressures are related to their mole fractions according to Dalton's Law of Partial Pressures. The fraction of the total pressure exerted by a gas is equal to its mole fraction in the mixture.

Let the mass of hydrogen ($H_2$) and ethane ($C_2{H}_6$) mixed in the container be equal, say w grams each.

The molar mass of hydrogen ($H_2$) is:
$M_{H_2}=2\text{g/mol}$

The molar mass of ethane ($C_2{H}_6$) is:
$M_{C_2{H}_6}=(2\times 12)+(6\times 1)=30\text{g/mol}$

Now, we calculate the number of moles (n) for each gas:
Moles of hydrogen, $n_{H_2}=\frac{w}{2}$
Moles of ethane, $n_{C_2{H}_6}=\frac{w}{30}$

The total number of moles in the mixture is:
$n_{\text{total}}=n_{H_2}+n_{C_2{H}_6}=\frac{w}{2}+\frac{w}{30}$
$n_{\text{total}}=\frac{15w+w}{30}=\frac{16w }{30}$

According to Dalton's Law of Partial Pressures, the fraction of the total pressure exerted by hydrogen is equal to its mole fraction:
$X_{H_2}=\frac{n_{H_2}}{n_{\text{total}}}$
$X_{H_2}=\frac{\frac{w}{2}}{\frac{16w}{30}}=\frac{w}{2}\times \frac{30}{16w}=\frac{15}{16}$

Therefore, the fraction of the total pressure exerted by hydrogen in the mixture is 15/16.