Please help with both part A and B below. Thank You!:) Part A If 0.200 mol of a

Question

Please help with both part A and B below. Thank You!:)

Part A If 0.200 mol of a nonvolatile nonelectrolyte are dissolved in 3.80 mol of water, what is the vapor pressure P, the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 °C of Express your answer with the appropriate units. Hints PHOValue Units Submit My Answers Give Up Solutions containing volatile solutes In solutions composed of two liquids (A and B), each liquid contributes to the total vapor pressure above the solution. The total vapor pressure is the sum of the partial pressures of the components: B1 B where Po and P3 are the vapor pressures of pureA andB, respectively. Part B A solution is composed of 1.90 mol cyclohexane (P, 97.6 torr) and 2.20 mol acetone (Pa = 229.5 torr) What is the total vapor pressure Ptotal above this solution? Express your answer with the appropriate units.

Explanation / Answer

A)

n(H2O),n1 = 3.8 mol

n(non electrolyte),n2 = 0.2 mol

Total number of mol = n1+n2

= 3.8 + 0.2

= 4 mol

we have below equation to be used:

Mole fraction of each components are

X(H2O) = n1/total mol

= 3.8/4

= 0.95

we have below equation to be used:

X(non electrolyte) = n2/total mol

= 0.2/4

= 0.05

According to Raoult’s law:

P = Po*X(solvent)

p = 23.8*0.95

p = 22.6 torr

Answer: 22.6 torr

B)

X1 = mol of cyclohexane / total mole

= 1.90 / (1.90 + 2.20)

= 0.463

X2 = mol of acetone / total mole

= 2.20 / (1.90 + 2.20)

= 0.537

P = Po1*X1 + Po2*X2

= 97.6*0.463 + 229.5*0.537

= 168 torr

Answer: 168 torr

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