Question

# Part A. If 0.680 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 mol of water,...

Part A. If 0.680 mol of a nonvolatile nonelectrolyte are dissolved in 3.90 mol of water, what is the vapor pressure PH2O of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 ∘C .

Part B. A solution is composed of 1.00 mol cyclohexane (P∘cy=97.6 torr) and 2.50 mol acetone (P∘ac=229.5 torr). What is the total vapor pressure Ptotal above this solution?

A)

n = 0.68 mol of nonvolatile

n = 3.9 mol water

P° = 23.8 torr

the

dP = x*P°

x = mol fraction of nonvolatile subs

x = mol S / (total mol)

total mol = 3.9+0.68 = 4.58

x = 0.68/4.58 = 0.14847

dP= x*P° = 0.14847*23.8 = 3.533586 torr

Pfinal = P° - dP = 23.8-3.533586 = 20.2664 torr

B)

n = 1

n = 2.5

this must be used with raoult law

P°mix = Xa*Pa° + Xb*Pb°

Xa = 1/(1+2.5) = 0.285714

Xb = 2.5/(1+2.5) = 0.7142

P°mix = 0.285714*97.6 + 0.7142*229.5 = 191.7945864 torr

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