What is the molality of benzene in a solution that contains $6.502 g$ $"6.502 g"$ of benzene in $162.7 g$ $"162.7 g"$ of chloroform/ ${CHCl}_{3}$ $"CHCl"_3$ ( $K_{f}$ $K_f$ for ${CHCl}_{3}$ $"CHCl"_3$ is ${4.68}^{\circ} C \cdot kg/mol$ $4.68^@ "C"cdot"kg/mol"$ and $T_{f}^{}$ $T_f^""$ of ${CHCl}_{3}$ $"CHCl"_3$ is ${63.5}^{\circ} C$ $63.5^@ "C"$ )?

Question

What is the molality of benzene in a solution that contains $6.502 g$ $"6.502 g"$ of benzene in $162.7 g$ $"162.7 g"$ of chloroform/ ${CHCl}_{3}$ $"CHCl"_3$ ( $K_{f}$ $K_f$ for ${CHCl}_{3}$ $"CHCl"_3$ is ${4.68}^{\circ} C \cdot kg/mol$ $4.68^@ "C"cdot"kg/mol"$ and $T_{f}^{}$ $T_f^""$ of ${CHCl}_{3}$ $"CHCl"_3$ is ${63.5}^{\circ} C$ $63.5^@ "C"$ )?

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Answer 1 · 2018-02-15T14:01:09

Molality is defined as moles of solute per $1 kg = 1000 g$ $"1 kg = 1000 g"$ of solvent. Convert benzene's mass to moles, then divide by the mass of the solvent in kilograms.

Explanation:

$6.0502 g \cdot \frac{1 mol}{78.114 g} = 0.083238 moles$ $"6.0502 g" * ("1 mol")/("78.114 g") = "0.083238 moles"$

So

$molality = \frac{0.083238 moles}{162.7 \cdot 10^{- 3} kg} = {0.51156 mol kg}^{- 1}$ $"molality" = "0.083238 moles"/(162.7 * 10^(-3) " kg") = "0.51156 mol kg"^(-1)$

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