Question

A solution of 6.54 g of a carbohydrate in 102.5 g of water has a density of 1.024 g/mL and an osmotic pressure of 4.61 atm at 20.0°C. Calculate the molar mass of the carbohydrate.

Answer 1

The molar mass of the carbohydrate : = 304.19 g/mol

Further explanation

Given

6.54 g carbohydrate

102.5 g of water

osmotic pressure of 4.61 atm

T = 20+273=293 K

Required

The molar mass

Solution

General formula:  

$\large {\boxed {\bold {\pi \: = \: M \: x \: R \: x \: T}}}$

π = osmosis pressure (atm)  

M = solution concentration (mol / l)  

R = constant = 0.08205 L atm mol-1 K-1  

T = Temperature (Kelvin)  

Find molarity(M) :

4.61 atm = M . 0.08205 x 293

M = 0.192 mol/L(mol solute per 1 liter solution)

Total mass of solution :

= 6.54 g + 102.5

= 109.04 g

Volume of solution :

= density x mass

= 1.024 g/ml x 109.04 g

= 111.66 ml

= 0.112 L

mol Carbohydrate (solute):

= M x V

= 0.192 x 0.112

= 0.0215 mol

Molar mass of Carbohydrate :

= mass : mol

= 6.54 : 0.0215

= 304.19 g/mol