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# A solution is prepared by dissolving 43 g of naphthalene in 117 g of benzene. Calculate the mole fractions of the two components of the solution.

Moles of benzene, $n_1=\large\frac{117}{78}$$=1.50mol Moles of naphthalenen_2=\large\frac{43}{128}$$=0.34mol$
Mole fraction of naphthalene, $X_2=\large\frac{0.34}{1.50+0.34}$$=0.815$
Mole fraction of benzene = 1 − $X_2$ = 1 − 0.185 = 0.815