Answer to Question #224801 in General Chemistry for sam

Question

1.At 35.0 oC, the vapour pressure of pure ethanol (C2H5OH) is 100.0 mm Hg and the vapour pressure of pure acetone (CH3COCH3) is 360.0 mm Hg.

a) A solution is formed from 25.0 g of ethanol and 15.0 g of acetone. Assume ideal behaviour and calculate the vapour pressure of each component above this solution.

b) Calculate the mole fraction of acetone in the vapour in equilibrium with the solution in part a).

c) Name a method that can used to separate these two components in the lab.

2. How many ice cubes (at 0.0 oC) would you have to add to a 100.0 mL drink initially at 25.0 oC in order to lower its temperature to 4.0 oC? Each ice cube weighs 5.00 g, and the density of water is 1.00 g mL-1. Assume no heat loss. You may find some or all of the following constants useful:

Specific heat capacity of solid ice = 2.06 J g-1 oC-1; ΔHfusion = 6.02 kJ mole-1

Specific heat capacity of liquid water = 4.184 J g-1 oC-1; ΔHvap = 40.7 kJ mole-1

Specific heat capacity of water vapor = 1.84 J g-1 oC-1;

Answer 1

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q1+q2=−q3(1)

, where

q1 - the heat absorbed by the solid at 0°C

q2 - the heat absorbed by the liquid at 0°C

q3 - the heat lost by the warmer water sample

The two equations that you will use are

q=m⋅c⋅ΔT

, where

q - heat absorbed/lost

m - the mass of the sample

c - the specific heat of water, equal to

4.18Jg°C

ΔT - the change in temperature, defined as final temperature minus initial temperature

and

q=n⋅ΔHfus

, where

q - heat absorbed

n - the number of moles of water

ΔHfus - the molar heat of fusion of water, equal to

6.01 kJ/mol

This is equivalent to

0.418⋅Tf+2.717⋅Tf=67.925−33.36

=34.565