Now suppose, instead, that 5.678 g of a volatile solute is dissolved in 150.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 20°C, a vapour pressure of 1.754 torr. Again, assume an ideal solution. If, at 20°C the vapour pressure of this solution is also 17.344 torr. Calculate the molar mass of this volatile solute.

59.9 g/mol is the molar mass for the solute

Explanation:

Lowering vapor pressure → ΔP = P° . Xm

P° → Vapor pressure of pure solvent

ΔP = P° - Vapor pressure of solution

Xm = Mole fraction of solute

17.54 Torr  - 17.344 Torr = 17.54 Torr . Xm

0.196 Torr / 17.54 Torr = Xm → 0.0112

These are the moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

We determine the moles of solvent → 150 g . 1mol/ 18 g = 8.33 moles

Now we can make this equation:

0.0112 = Moles of solute / Moles of solute + 8.33 mol

0.0112 Moles of solute + 0.0933 = Moles of solute

0.0933 = Moles of solute - 0.0112 Moles of solute

0.0933 = 0.9888 moles of solute → 0.0933 / 0.9888 = 0.0947 moles

Finally we can determine the molar mass (mol/g)

5.678 g / 0.0947 mol = 59.9 g/mol

608 L

Explanation:

Using Boyle's law equation as follows:

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

According to the information provided in this question;

P1 = 1.00atm

P2 = 2.8atm

V1 = ?

V2 = 217L

Using P1V1 = P2V2

V1 = P2V2/P1

V1 = 2.8 × 217/1

V1 = 607.6/1

V1 = 607.6

The initial volume is 608 L