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.

Answer 2

Considering the definition of mole fraction and molality:

• the mole fraction of benzene is 0.077.
• the molality of benzene is 0.9908 $\frac{moles}{kg}$.

You know that:

• Mass of benzene = 6.24 grams
• Mass of cyclohexane= 80.74 grams
• Molar mass of benzene= 78.11 g/mole
• Molar mass of cyclohexane= 84.16 g/mole

Mole fraction

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

Being the molar mass and the mass in the solution of each compound, the number of moles of each compound can be calculated as:

• Benzene: $\frac{mass of benzene}{molar mass of benzene} =\frac{6.24 grams}{78.11 \frac{grams}{mole} } = 0.08 moles$
• Cyclohexane:$\frac{mass of cyclohexane}{molar mass of ciclohexane} =\frac{80.74 grams}{84.16\frac{grams}{mole} } = 0.96 moles$

So, the total moles of the solution can be calculated as:

Total moles = 0.08 moles + 0.96 moles = 1.04 moles

Finally, the mole fraction of benzene can be calculated as follow:

$\frac{number moles of benzene}{total moles} =\frac{0.08 moles}{1.04 moles} = 0.077$

Finally, the mole fraction of benzene is 0.077.

Molality

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

$Molality=\frac{number of moles of solute}{kilograms of solvent}$

In this case, you know:

• number of moles of solute= 0.08 moles
• Mass of solvent = 80.74 g = 0.08074 kg (being 1000 g=1 kg)  

Replacing:

$Molality benzene=\frac{0.08 moles}{0.08074 kg}$

Molality benzene= 0.9908 $\frac{moles}{kg}$

Finally, the molality of benzene is 0.9908 $\frac{moles}{kg}$.

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