Question

question 24 of 36 >
imagine two solutions with the same concentration and the same boiling point, but one has ethanol as the solvent and the other has carbon tetrachloride as the solvent. determine that molal concentration, $m$ (or $b$), and boiling point, $t_b$.

solvent	ethanol	carbon tetrachloride
$k_b$ (°c/m)	1.22	5.03

$m = \boldsymbol{\square}$ $m$
$t_b = \boldsymbol{\square}$ °c

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Explanation:

Step1: Set boiling point equality

Let $\Delta T_{b,ethanol} = T_b - T_{b,ethanol}^\circ$ and $\Delta T_{b,CCl_4} = T_b - T_{b,CCl_4}^\circ$. Since boiling points are equal, use $\Delta T_b = K_b \cdot m$:
$$T_{b,ethanol}^\circ + K_{b,ethanol} \cdot m = T_{b,CCl_4}^\circ + K_{b,CCl_4} \cdot m$$

Step2: Rearrange to solve for $m$

Isolate terms with $m$ on one side:
$$K_{b,ethanol} \cdot m - K_{b,CCl_4} \cdot m = T_{b,CCl_4}^\circ - T_{b,ethanol}^\circ$$
$$m(K_{b,ethanol} - K_{b,CCl_4}) = T_{b,CCl_4}^\circ - T_{b,ethanol}^\circ$$
$$m = \frac{T_{b,CCl_4}^\circ - T_{b,ethanol}^\circ}{K_{b,ethanol} - K_{b,CCl_4}}$$

Step3: Substitute given values

Plug in $T_{b,ethanol}^\circ=78.4^\circ\text{C}$, $T_{b,CCl_4}^\circ=76.8^\circ\text{C}$, $K_{b,ethanol}=1.22^\circ\text{C/m}$, $K_{b,CCl_4}=5.03^\circ\text{C/m}$:
$$m = \frac{76.8 - 78.4}{1.22 - 5.03} = \frac{-1.6}{-3.81} \approx 0.420$$

Step4: Calculate final boiling point $T_b$

Use ethanol's boiling point formula:
$$T_b = T_{b,ethanol}^\circ + K_{b,ethanol} \cdot m$$
$$T_b = 78.4 + (1.22 \times 0.420) \approx 78.4 + 0.512 = 78.912$$

Answer:

$m = 0.420$ m
$T_b = 78.9$ °C