Question

A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calculate the mole fraction and molality of benzene in this solution.

Answer 1

Answer:

$x_B=0.0769$

$m=0.990m$

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

$x_B=\frac{n_B}{n_B+n_C}$

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

$n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol$

Thus, we compute the mole fraction:

$x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769$

Next, for the molality, we define it as:

$m=\frac{n_B}{m_C}$

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

$m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg$

Or just 0.990 m in molal units (mol/kg).

Best regards.