Question

A solution containing 8.6 g urea in 1L was found to be isotonic with 5% (weight/volume) of an organic non-volatile solute. The molecular weight in g/mol of later is:
A.348.83
B.34.89
C.3489
D.861.2

Answer 1

We know that two solutions when isotonic indicates that concentration of both the solutions is equal. Here, moles of solute and solvent of two solutions are given. So, we have to equate concentration of both the solutions to calculate molecular weight of the organic non volatile solute. The formula of concentration we have to use here is, ${\rm{Concentration}} = \dfrac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{solution}}\,{\rm{in}}\,{\rm{Litre}}}}$

Complete step by step answer:
Let’s first understand why two isotonic solutions have equal concentration. By the term isotonic we understand that both the solutions equal osmotic pressure. The formula of osmotic pressure is $\pi = CRT$ , where, $\pi$ is the osmotic pressure, C is concentration, R is gas constant and T is temperature. Solution (1) and (2) is isotonic only when

${\pi _1} = {\pi _1}$

$\Rightarrow {C_1}RT = {C_2}RT$

$\Rightarrow {C_1} = {C_2}$ ………………….…… (1)

So, it is proved that when two solutions are isotonic, they have equal concentration.

Now, come to the question. The mass of urea is 8.6 g and volume of the solution is 1 L. So,
Mass of solute =8.6 g and volume of solution =1 L.

Now, we calculate the concentration of urea solution. For this, we need moles of urea and volume of volume. For calculation of number of moles we use the formula,

${\rm{Number}}\,{\rm{of moles}} = \dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{solute}}}}$

So, for urea $\left( {{\rm{CO}}{{\left( {{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)}_2}} \right)$
molar mass is=$= 12 + 16 + 2 \times 14 + 4 \times 1 = 60\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$.

$\Rightarrow {\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of urea}} = \dfrac{{8.6\,{\rm{g}}}}{{60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}$

Now, we calculate the concentration of urea solution.

${C_1} = \dfrac{{\dfrac{{8.6\,{\rm{g}}}}{{60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}}}}{{1\,L}} \Rightarrow \dfrac{{8.6\,{\rm{g}}}}{{60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} \times 1\,{\rm{L}}}}$

Similarly, we have to calculate the concentration of solution of non-volatile solute. Given that, the solution is 5% (weight/volume). So, the mass of solute is 5 g and the volume of solution is 100 ml (0.1 L).

To calculate moles of the non volatile solute, we take the molar mass as M.

So, ${\rm{Moles}}\,{\rm{of}}\,{\rm{non - volatile}}\,{\rm{solute}} = \dfrac{5}{M}$

Now, we calculate the concentration of the solution. Volume should be converted to litre.

${C_2} = \dfrac{{\dfrac{{5\,}}{{\rm{M}}}}}{{\dfrac{{{\rm{100}}\,}}{{1000\,}}}}$

$\Rightarrow {C_2} = \dfrac{{5\,}}{{\,{\rm{M}}}} \times \dfrac{{1000}}{{100}}$

Now, we have to put the value of ${C_1}$ and ${C_2}$ in equation (1)

$\dfrac{{8.6\,}}{{60\, \times 1\,}} = \dfrac{{5\,\,}}{{\,{\rm{M}}}} \times \dfrac{{1000}}{{100\,}}$

$\Rightarrow {\rm{M}} = \dfrac{{5\, \times 10 \times 60}}{{8.6\,}}$

$\Rightarrow M = 348.83\,\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$

Therefore, molar mass of non-volatile solute is $348.83\,\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$.

So, the correct answer is Option A .

Note:
Remember that, minimum pressure that is required by a solution for flowing of solvent through a semipermeable membrane is termed as osmotic pressure. It is represented by the symbol $\pi$. It is one of the colligative properties.