Answer to Question #139184 in Organic Chemistry for Emily

Question #139184
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).
1
Expert's answer
2020-10-19T14:27:41-0400

For CCl4:

Kf=29.8CmK_f=29.8 \frac{C^\circ}{m} , Tf=22.9CT_f=-22.9 C^\circ

Kb=5.03CmK_b= 5.03 \frac{C^\circ}{m} , Tb0=76.8CT_b^0=76.8 C^\circ


1) Find molality of a solution

20.0g(C10H8)×1mole(C10H8)128.17g(C10H8)=0.156mol(C10H8)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)


mmolarity=mol(C10H8)kg(CCl4)=0.156mol(C10H8)0.1607kg(CCl4)=0.917molalmolarity = \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 TfrsolutionT_{fr}{solution}

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

ΔTfr=TsolventTsolution\Delta T_{fr}= T_{solvent}- T_{solution}

Tsolution=TsolventΔTfr=22.928.94=51.84CT_{solution} = T_{solvent} - \Delta T_{fr} = -22.9-28.94 = -51.84 C^\circ


3) Find Tb(solution)T_b(solution)

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

Tsolution=Tsolvent+ΔTb.p.=76.8+4.88=81.68CT_{solution}= T_{solvent} +\Delta T_{b.p.}=76.8+4.88=81.68 C^\circ


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

π=iMRTπ=iMRT

i=1i=1

M=molarity=moles(C10H8)volumesolutionM =molarity= \frac{moles (C_{10}H_8)}{volume_{solution}}

V=md=20.0+160.71.253=144.2mlV=\frac{m}{d}= \frac{20.0+160.7}{1.253}=144.2 ml


M=molarity=0.156moles144.2×103=1.082MM= molarity = \frac{0.156 moles}{144.2\times 10^{-3}}= 1.082 M

π=1×1.082×0.08206×298=26.5atmπ=1\times 1.082\times 0.08206\times 298= 26.5 atm


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