A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 258 torr . Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature. What is the mole fraction of hexane? (Assume ideal behavior.)

The mole fraction of hexane is 0.61

Explanation:

Raoult's law allows calculating the vapor pressure of a substance when it is forming part of an ideal solution, knowing its vapor pressure when it is pure (at the same temperature) and the composition of the ideal solution in terms of mole fraction.

Pi=xi*Pi°

where Pi is the partial pressure of component i in the solution, Pi ° is the vapor pressure of component i pure and xi is the mole fraction of component i in the liquid phase.

In a solution, once the components have reached chemical equilibrium, the total vapor pressure is:

Psolution=P1 + P2=x1*P1° +x2*P2°

In this case:

Psolution= xpentane*Ppentane° +xhexane*Phexane°

Being:

and replacing:

258 torr= (1-xhexane)* 425 torr + xhexane* 151 torr

you get:

258 torr= 425 torr - 425 torr*xhexane + xhexane* 151 torr

258 torr - 425 torr= - 425 torr*xhexane + xhexane* 151 torr

-167 torr= -274 torr*xhexane

xhexane= 0.61

The mole fraction of hexane is 0.61

chemistry sucks a**

explanation:

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