Question

 

1. Determine the mole fraction composition for the following mixture.

The mixture solution is composed of two volatile components:
   Solute: pentane (72.0 g/mol)
   Solvent: hexane (86.0 g/mol)

The vapor pressures are the following:
   Solution: 258 torr
   Solute Pentane: 425 torr
   Solvent Hexane: 151 torr

Assume ideal behavior and that the mixture is at room temperature

The answer is
Xpentane = 0.391
Xhexane = 0.609

2. A 35.0-g sample of hemogoblin is dissolved in enough water to make 1.00L of solution. The osmotic pressure (pi) was measured to be 10.0 mmHg at
25°C

Answer: Calculate the molar mass of hemoglobin
6.49*10^4 g/mol

use Osmotic formula
Pi = MRT

Answer 1

2. A 35.0-g sample of hemogoblin is dissolved in enough water to make 1.00L of solution. The osmotic pressure (pi) was measured to be 10.0 mmHg at
25°C
Answer: Calculate the molar mass of hemoglobin
6.49*10^4 g/mol
use Osmotic formula

Solution: Examine the osmosis equation:
Pi = M R T
for what is given and what is not given in the problem.
We have:
1) osmotic pressure (Pi) = 10 mmHg = 0.01315 atm
2) molarity (M) = X
3) a constant (R) = 0.08206 L atm mol¯¹ K¯1
4) temperature (T) = 25 °C = 298 K
. let's calculate it. Notice I've included a conversion of mmHg to atm.
(0.01315) = x * (0.08206 L atm mol¯1 K¯1) (298 K)

x = 5.377 X 10^-4 mol L¯1
Molarity of hemoglobin =5.377 X 10^-4 mol L¯1

Molarity = weight of hemoglobin / Molar mass of hemoglobin

Molar mass of hemoglobin= weight of hemoglobin / Molarity
= 35.0 / 5.377 X 10^-4 mol L¯1
= 6.49 X 10⁴ gm/mole- Required Answer