Question

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g$mo{{l}^{-1}}$ ) in the same solvent. If the densities of both the solutions are assumed to be equal to $1.0\,gc{{m}^{-3}}$, molar mass of the substance will be:
A.90.0$gmo{{l}^{-1}}$
B. 115.0$gmo{{l}^{-1}}$
C. 105.0 $gmo{{l}^{-1}}$
D. 210.0$gmo{{l}^{-1}}$

Answer 1

Hint: The movement of particles between concentration gradients is called osmosis. When extra pressure is applied on the solution to stop osmosis, then it is called osmotic pressure. Isotonic solutions have the same osmotic pressures. Osmotic pressure depends on the mass of solute so it is a colligative property.
Formula used:
$\pi =CRT$ (where C is concentration, R gas constant and T is temperature)

Complete answer:
We have been given two isotonic solutions with their concentrations as 5.25% and 1.5% solution of urea, we know the molar mass of urea as 60 g/mol and the densities of both are equal as $1.0\,gc{{m}^{-3}}$.
As the solutions are isotonic so they have same osmotic pressure this means that ${{\pi }_{1}}={{\pi }_{2}}$
That means ${{C}_{1}}RT={{C}_{2}}RT$
Therefore, concentration converted to moles per liter as R and T being cancelled, the equation becomes:
$\dfrac{5.25}{{{M}_{1}}}\times \dfrac{1000}{100}=\dfrac{1.5}{60}\times \dfrac{1000}{100}$ where, ${{M}_{1}}$ is taken as the molar mass of liquid 1, while liquid 2 is urea with mass 60 g/mol.
So, Molar mass of liquid 1 = $\dfrac{5.25\times 60}{1.5}$
Molar mass = 210$gmo{{l}^{-1}}$
So, the molar mass is 210.0$gmo{{l}^{-1}}$

Therefore option D is correct.

Note:
The concentration is expressed as number of moles upon volume $\dfrac{Number\,of\,moles}{Volume}$, where number of moles are the weight of solute upon molar mass of solute, weight of solute is calculated from percentage as 5.25% = 5.25 g of solute dissolved in 100 mL solvent. So, the conversion factor uses multiplication by 1000 that indicates the volume.