Question

A solution containing 10.2 g glycerine per litres is isotonic with a \[2\% \] solution of glucose. What is the molecular mass of glycerine?
A.91.8 g
B.1198 g
C.83.9 g
D.890.3 g

Answer 1

Hint: If two solutions are isotonic, they have the same osmotic pressure across the semipermeable membrane. And we know Osmotic Pressure , $\pi = \dfrac{n}{v}RT$ (and $n = \dfrac{w}{m}$ )so osmotic pressure of glycerine solution will be equal to osmotic pressure of glucose solution i.e. $\pi Glycerine = \pi Glucose$ . Putting the values in place, we will find the molecular mass of glycerine.

Complete step by step answer:
Given in the question,
Total glycerine in the solution – 10.2 g
Glucose solution \[ - 2\% \] of glucose in the solution
Since, both solutions are isotonic
We know, Osmotic Pressure, $\pi = \dfrac{n}{v}RT = \dfrac{W}{{MV}}RT$
Where, V = Volume of Solution
R = Gas Constant
T = Temperature
W = Weight of solute
M = Molecular weight of solute
 $\pi Glycine = \pi Glu\cos e$
$\Rightarrow$ $\dfrac{{n1}}{{V1}}RT = \dfrac{{n2}}{{V2}}RT$
 \[2\% \] of solution means 2 g glucose in 100 g solution,
$\Rightarrow$ $\dfrac{{Wglycerine}}{{Mglycerine}} \times \dfrac{1}{{V1}} = \dfrac{{Mglu\cos e}}{{Wglu\cos e}} \times \dfrac{1}{{V2}}$
Molecular mass of Glucose $({C_6}{H_{12}}{O_6}) = (12 \times 6) + (1 \times 12) + (16 \times 6) = 180$
(Mass of C atom = 12, mass of Hydrogen = 1 and mass of Oxygen = 16)

$\Rightarrow$ $\dfrac{{10.2}}{{Mglycerine}} \times \dfrac{1}{{1000}} = \dfrac{2}{{180}} \times \dfrac{1}{{100}}$
 ${M_{glycerine}}$ =91.8g
The molecular mass of glycerine is 91.8g.

Therefore, the correct answer is option (A).

Note: The minimum pressure that is applied to a solution to halt the flow of solvent molecules through a semipermeable membrane (osmosis) is known as the osmotic pressure. The osmotic pressure can be calculated by the equation, $\pi = iCRT$ . The semi permeable membrane allows only the movement of solvent molecules through it as they cannot pass through the membrane but the solute particles cannot pass through it. When the two solutions are isotonic, they have the same concentration in moles/litre.