Question

In a laboratory experiment, a student found that a 158-mL aqueous solution containing 2.589 g of...

In a laboratory experiment, a student found that a 158-mL aqueous solution containing 2.589 g of a compound had an osmotic pressure of 13.5 mm Hg at 298 K. The compound was also found to be nonvolatile and a nonelectrolyte. What is the molar mass of this compound?

The nonvolatile, nonelectrolyte aspirin, C9H8O4 (180.1 g/mol), is soluble in chloroform CHCl3.
Calculate the osmotic pressure (in atm) generated when 14.4 grams of aspirin are dissolved in 185 mL of a chloroform solution at 298 K.

Homework Answers

Answer #1

1)

P= 13.5 mm Hg

= (13.5/760) atm

= 1.776*10^-2 atm

T = 298 K

use:

P = C*R*T

1.776*10^-2 = C*0.08206*298.0

C =7.264*10^-4 M

volume , V = 1.58*10^2 mL

= 0.158 L

use:

number of mol,

n = Molarity * Volume

= 7.264*10^-4*0.158

= 1.148*10^-4 mol

mass(solute)= 2.589 g

use:

number of mol = mass / molar mass

1.148*10^-4 mol = (2.589 g)/molar mass

molar mass = 2.256*10^4 g/mol

Answer: 2.26*10^4 g/mol

2)

  

mass, m = 14.4 g

use:

number of mol,

n = mass/molar mass

=(14.4 g)/(180.1 g/mol)

= 7.996*10^-2 mol

volume , V = 1.85*10^2 mL

= 0.185 L

use:

Molarity,

M = number of mol / volume in L

= 7.996*10^-2/0.185

= 0.4322 M

T = 298 K

use:

P = C*R*T

P = 0.4322*0.08206*298.0

P =10.57 atm

Answer: 10.6 atm

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