Question

A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1.00 l of solution. the osmotic pressure of this solution is 0.750 atm at 25.0^\circ c 25.0 ∘

c. what is the molecular weight (g/mol) of the unknown solute?

Answer 1

Osmotic pressure can be calculated using the following equation:

`\prod =C* R* T`

Here,

C representated concentration

R represented gas constant

T represented temperature

`\prod` representated osmotic pressure

R=0.0821 atm L mol ⁻¹

T = 25 + 273 = 298 K

C=
`(Given mass of the substance)/(Molar mass of the substance* Volume)`

Given mass of the substance= 6.00 g

Volume= 1 L

`\prod` =0.75 atm

Putting all the values in the equation:

`0.750 =(6)/(Molar mass of the substance* 1)* 0.0821* 298`

Molar mass of the substance= 195 g mol⁻¹.