Given that the boiling points of ethanol and diethyl ether are 78.1°c and 34.6°c, respectively, which should have the higher vapor pressure at 20.0°c?

- Diethyl ether would have higher vapor pressure at 20.0 degree C.

Explanations:- Boiling point is the temperature at which the vapor pressure equals to the atmospheric pressure.

Looking at the given vapor pressure values, ethanol needs to be heat up at 78.1 degree C to makes it's vapor pressure equals to atmospheric pressure where as diethyl ether only needs to be heat up at 34.6 degree C.

Hence, lower is the boiling point, higher is the vapor pressure.

Now, 20.0 degree C is more closer to 34.6 degree C . it means the temperature difference is less for diethyl ether as compared to ethanol.

So, as the boiling point of diethyl ether is low, the vapor pressure of it would be higher at 20.0 degree C.

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