Asolution is made by mixing 33.0 ml of ethanol, c2h6o and 67.0 ml of water. assuming ideal behavior, what is the vapor pressure of the solution at 20 degree c? ethanol= .789 g/ml 43.9 torr. water= .998 g/ml 17.5 torr

The vapor pressure of the solution is 20.757 torr.

According to Raoult's law, the vapor pressure can be given by:

Vapor pressure = Mole fraction Partial pressure

Density =

Mass = Density Volume

Mass of ethanol = 0.789 g/ml 33 ml

Mass of ethanol = 26.037 grams

Mass of water = 0.998 g/ml 67

Mass of water = 66.866 grams

Moles =

Moles of ethanol =

Moles of ethanol = 0.566 mol

Moles of Water =

Moles of water = 3.714 mol

Total moles of solution = Moles of water + Moles of ethanol

Total moles of solution = 3.714 mol + 0.566 mol

Total moles of solution = 4.28 mol

Mole fraction =

Mole fraction of ethanol =

Mole fraction of ethanol = 0.13

Mole fraction of water =

Mole fraction of water = 0.86

Vapor pressure of ethanol = 0.13 43.9

Vapor pressure of ethanol = 5.707 torr

Vapor pressure of water = 0.86 17.5

Vapor pressure of water = 15.05 torr

The vapor pressure of solution = Vapor pressure of ethanol + Vapor pressure of water

The vapor pressure of solution = 5.707 + 15.05 torr

The vapor pressure of solution = 20.757 torr.

The vapor pressure of the solution is 20.757 torr.

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol