5. Raoult\'s Law. A solution contains a mixture of pentane and hexane at 23 °C.

Question

5. Raoult's Law. A solution contains a mixture of pentane and hexane at 23 °C. The solution has a vapor pressure of 247 torr. Pure pentane and pure hexane have vapor pressures of 425 torr and 151 torr, respectively at 23 °C. What is the mole fraction of the mixture? Assume Ideal behavior. 6. Boiling Point Elevation. A non-electrolyte contains 38.7% C, 9.7% H, and 51.6% O. 15.5 g of this solute is dissolved in 100.0 g of water, and the resulting solution boils at 101.30 °C. What is the molecular formula of the solute? (K,for water is 0.512 °Clm

Explanation / Answer

1. total pressure = pure pressure of pentane * mole fraction of pentane + pure pressure of hexane * mole fraction of hexane

P(total)=P(hexane )* x(hexane) +P(pentane)*x(pentane) x(hexane) +x(pentane) =1

x(hexane)=1-x(pentane)

putting in the equation

P(total)=P(hexane)*(1-x(pentane))+P(pentane)*x(pentane)

P(total)=P(hexane)+(P(pentane)-P(hexane))*x(pentane )

now putting the values

247=151+(425-151)*x(pentane)

solving we get x(pentane)= 0.35

x(hexne)=1-0.35=0.65

2. delta T = Kb* m * i

where m is molarity

i =1 (vant hoff factor) for substance which do not dissociate in compound

putting the values , boiling point of water is 100

delta T = 101.30-100=1.30

1.30=0.512*m*1

m= 2.53 mol/l

now C-38.7% ----divided by atomic mass =38.7/12=3.225

H-9.7%------divided by atomic mass=9.7/1=9.7

O-15.5%-----divided by atomic mass =51.6/16=3.225

now divide each number by smaller one that is 3.225

ratio of carbon:hydrogen:oxygen=1:3:1

our empirical formula = CH3O empirical formula mass= 31

suppose density of water = 1 so volume = 100ml

now molarity we have calculated above is =2.53

means 2.53 mole in 1000 ml

now in 100 ml 0.253 mole by unitary method

0.253 corresponds 15.5 g so 1 mole correspond=15.5/0.253=61.26

now n = 61.26/31==1.9 or 2

so our formula of compound =C2H6O2

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