Question

0.1 g of a substance dissolved in 22g of benzene lowers the freezing point of benzene by 0.2 degree celsius . calculate the molecular mass of the substances (Kf= 12 degree celsius mol

Answer 1

#82745 Chemistry, Other

0.1 g of a substance dissolved in 22 g of benzene lowers the freezing point of benzene by 0.2 degree celsius. Calculate the molecular mass of the substances ( $K_{\mathrm{f}} = 12$ degree / celsius mol).

Answer:

$$\begin{array}{l} \Delta T _ {\mathrm {f}} = K _ {\mathrm {f}} \times C m \\ 0. 2 = 1 2 \times m \\ C m = 0. 2 / 1 2 = 0. 0 1 7 \text { molal} \\ C m = n _ {\text {(s o l u t e)}} / m _ {\text {(s o l v e n t)}} \quad n _ {\text {(s o l u t e)}} = n _ {\text {(s o l u t e)}} \times m _ {\text {(s o l v e n t)}} \\ n _ {\text {(s o l u t e)}} = 0. 0 1 7 \times 2 2 / 1 0 0 0 = 0. 0 0 0 4 \text { mol} \\ n = m / M \quad M = m / n \\ M _ {\text {(s u b s t a n c e)}} = 0. 1 / 0. 0 0 0 4 = 2 5 0 \mathrm {g} / \mathrm {m o l} \\ \end{array}$$

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