Question

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 *P*total above this
solution?

Answer #1

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

Part A
If 0.560 mol of a nonvolatile nonelectrolyte are
dissolved in 3.40 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 1.90 mol cyclohexane
(P∘cy=97.6 torr) and 2.80 mol acetone (P∘ac=229.5
torr). What is the total vapor pressure Ptotal above this
solution?

As you saw in Part B, the vapor above the cyclohexane-acetone
solution is composed of both cyclohexane vapor and acetone vapor.
What mole fraction of the vapor above the solution, \(X_{\rm
cy}(\rm vapor)\), is cyclohexane?
Part B:
A solution is composed of 1.60 {\rm mol} cyclohexane (P_{\rm
cy}^\circ=97.6~ \rm torr ) and 2.40 {\rm mol} acetone (P_{\rm
ac}^\circ=229.5~ \rm torr ). What is the total vapor pressure
P_{\rm total} above this solution?
Mole fraction of Cyclohexane = 1.60/(1.60+2.40) =
0.40...

A solution is composed of 1.50 mol cyclohexane
(P∘cy=13.0 kPa) and 2.10 mol acetone
(P∘ac=30.6 kPa). What is the total vapor
pressure Ptotal above this solution?
Express your answer with the appropriate units.

A solution is prepared by mixing 3.46 mol of acetone with 1.45
mol of cyclohexane at 30°C. Calculate the χacetone and χcyclohexane
in the vapor above the solution.
P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr.

1-Sugar is dissolved in hot water at 45oC until the mole
fraction of water decreases to 0.929. The vapor pressure of pure
water at this temperature is 72 torr. What is the vapor pressure
(in torr) of water for this solution?
2-54.6 g of the nonvolatile solute urea (M.W. = 60.06 g/mol) is
dissolved in 316 g of water at 60oC. The vapor pressure of pure
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nonelectrolyte, in 309 g water. (The vapor pressure of pure water
is 23.8 torr at 25°C and 71.9 torr at 45°C.)
Calculate the vapor pressure of this solution at 25°C.
Vapor pressure = torr
Calculate the vapor pressure of this solution at 45°C.
Vapor pressure = torr

22.0 grams of a nonvolatile solute are added to 60.0 grams of
water. The vapor pressure above the solution is 21.2 Torr. Pure
water has a vapor pressure of 23.8 at this temperature. What is the
molecular weight of the solute?

One mole of a nonelectrolyte was dissolved in 250 g of water and
the solution heated to 100C. Given that the vapor pressure of water
at 100C is 760 torr, what is the vapor pressure of the solution at
this temperature. Show your calculations please!

a. The nonvolatile, nonelectrolyte chlorophyll,
C55H72MgN4O5
(893.50 g/mol), is soluble in
ethanol
CH3CH2OH.
Calculate the osmotic pressure generated when 10.4
grams of chlorophyll are dissolved in
184 ml of a ethanol solution at
298 K.
The molarity of the solution is _____ M.
The osmotic pressure of the solution is _______atmospheres.
b. The nonvolatile, nonelectrolyte chlorophyll
, C55H72MgN4O5
(893.5 g/mol), is soluble in diethyl
ether
CH3CH2OCH2CH3.
How many grams of chlorophyll are needed to
generate an osmotic pressure of 1.21 atm...

A solution is prepared by adding 1.50 mol glucose, which is not
volatile, to 3.50 mol water. What is the vapor pressure of this
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