Answer to Question #139184 in Organic Chemistry for Emily

Question

A 20.0 g sample of naphthalene, C10H8, is dissolved in 160.7 g of carbon tetrachloride. Calculate the melting point,
boiling point, and osmotic pressure (at 25 ºC) of the resulting solution. (Assume a density of 1.253 g/mL for the
solution).

Answer 1

For CCl4:

$K_f=29.8 \frac{C^\circ}{m}$ , $T_f=-22.9 C^\circ$

$K_b= 5.03 \frac{C^\circ}{m}$ , $T_b^0=76.8 C^\circ$

1) Find molality of a solution

$20.0 g (C_{10}H_8) \times \frac{1 mole (C_{10}H_8)}{128.17 g(C_{10}H_8)}=0.156 mol (C_{10}H_8)$

m$molarity = \frac{mol (C_{10}H_8)}{kg(CCl_4)}=\frac{0.156 mol (C_{10}H_8)}{0.1607 kg(CCl_4)}= 0.917 molal$

2) Find $T_{fr}{solution}$

$\Delta T_{fr}= K_f\times m = 29.8\times 0.917 = 28.94C^\circ$

$\Delta T_{fr}= T_{solvent}- T_{solution}$

$T_{solution} = T_{solvent} - \Delta T_{fr} = -22.9-28.94 = -51.84 C^\circ$

3) Find $T_b(solution)$

$\Delta T_{b.p.}= K_b\times m = 5.03\times 0.971 = 4.88 C^\circ$

$T_{solution}= T_{solvent} +\Delta T_{b.p.}=76.8+4.88=81.68 C^\circ$

4) Find osmotic pressure (at $25C^\circ)$

$π=iMRT$

$i=1$

$M =molarity= \frac{moles (C_{10}H_8)}{volume_{solution}}$

$V=\frac{m}{d}= \frac{20.0+160.7}{1.253}=144.2 ml$

$M= molarity = \frac{0.156 moles}{144.2\times 10^{-3}}= 1.082 M$

$π=1\times 1.082\times 0.08206\times 298= 26.5 atm$