Question

A solution containing 10.2 g per glycerin per litre of a solution is found to be isotonic with $2%$ solution of glucose, molar mass = 180. Calculate the molar mass of glycerin.
(A) $97.89\,g/mol$
(B) $183.24\,g/mol$
(C) $91.8\,g/mol$
(D) None of these

Answer 1

Hint: In the isotonic solution, since the concentration of the two solutions is the same, there is no flow of solution through the membrane as the concentration gradient is zero. Thus, the pressure exerted by the solutions are equal.

Complete step by step solution:
It is given that the two solutions of glucose and glycerine are isotonic solutions. That is, they have the same concentration and same osmotic pressure.
Then, we have, the Van’t Hoff law, which relates the osmotic pressure to the concentration of the solution, given as,
$pV=nRT$
where $\pi $ is the osmotic pressure (in atm), V is the volume of solution (in litres), n is the moles of solute, R is the gas constant, and T is the absolute temperature.
So, the osmotic pressure of glycerine, with 10.2 g of glycerine in 1 litres of solution, will be ${{\pi }_{glycerine}}\times 1=\dfrac{10.2}{M}\times RT$ -------- (a)
where, moles of glycerine, $n=\dfrac{mass}{molar\,mass}=\dfrac{10.2}{M}$
Similarly, the osmotic pressure of glucose solution with 2% concentration, that is, weight by volume percentage concentration of 2g glucose in 100ml (0.1L) of solution and molar mass=180 g/mol, will be,
${{\pi }_{glu\cos e}}\times 0.1=\dfrac{2RT}{180}$ -------- (b)
Equating equation (a) and (b) as osmotic pressure of both the solutions are equal, then, we have,
$\dfrac{10.2\,RT}{M}=\dfrac{2RT}{180\times 0.1}$
\[M=\dfrac{10.2\,RT}{1}\times \dfrac{180\times 0.1}{2RT}\]
$M=91.8\,gmo{{l}^{-1}}$

Therefore, the molar mass of the glycerine is option (C)- $91.8\,g/mol$.

Note: Even though the concentration and the osmotic pressure of the two solutions are the same, the volume is not equal and the process of flow of the ions from the higher concentration to the lower concentration is called osmosis.