Question

An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient water to form 250 mL of solution. The solution has an osmotic pressure of 1.2 atm at 25 ºC. What is the molar mass (g/mol) of the compound?

Answer 1

Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol