Question

A solution of cane sugar containing 18 g L^-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution.

(Molar mass of cane sugar = 342, R = 0.082 lit atm mol^-1K^-1)

Answer 1

Given : 

Amount of cane sugar = W = 18 g L^-1 

Osmotic pressure = π = 1.25 atm 

Molar mass of cane sugar = M = 342 g mol^-1

Temperature = T = ? 

Number of moles of cane sugar

= \(\frac{W}{M}\)

= \(\frac{18}{342}\)

= 0.05263 mol lit^-1

∴ Concentration of solution = C

= \(\frac{n}{V}\)

= \(\frac{0.05263}{1}\)

= 0.05263 mol lit^-1

π = CRT

∴ T = \(\frac{\pi}{CR}\)

= \(\frac{1.25}{0.05236\times 0.08206}\)

= 289.4 K

∴ Temperature of solution = 289.4 K