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Question

Calculate the mole fraction of benzene in solution containing $$30\%$$ by mass in carbon tetrachloride.


Solution

$$Cl_4 (n) = \dfrac{70}{154} = 0.455$$

$$nC_6H_6 = \dfrac{nC_6H_6}{nC_6H_6 + nCCl_4}$$

$$= \dfrac{0.385}{0.385 + 0.455}$$

$$= \dfrac{0.385}{0.84}$$

= 0.458

$$nCCl_4 = 1 - 0.458$$

= 0.542

Chemistry

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