Question

$5$ % solution of cane sugar is isotonic with $0.877$ % of X. The molecular weight of substance X is
(A)$126.98$
(B)$119.96$
(C)$95.5$
(D)$59.98$

Answer 1

Isotonic solutions are those which have the same osmotic pressure. If the osmotic pressures are equal at the same temperature then the concentration should be equal. So osmotic pressures need to be equated to find the required entity.

Complete step by step solution: We know that isotonic solutions are those whose osmotic pressures are equal, the temperatures being constant
The formula for osmotic pressure is given as follows
$\pi = CRT$ which is for non-electrolytes
Where $\pi$ is the osmotic pressure
C is the concentration which is in molarity
T is the temperature
R is the universal gas constant
Now coming to the question we equate the osmotic pressures of both the solutions.
$\pi = CRT$
${\pi _1} = {\pi _2}$
${C_1}RT = {C_2}RT$
$\Rightarrow {C_1} = {C_2}$
Now let the molecular weight of X be M $g\;mo{l^{ - 1}}$
The molecular weight of cane sugar$= 342\;g\;mo{l^{ - 1}}$
Let the osmotic pressure of cane sugar be ${\pi _c}$ and ${\pi _x}$ respectively
Equating both the osmotic pressures we get
${C_C}RT = {C_x}RT$
Cancelling RT from both sides as they are constant, we get
${C_C} = {C_x}$
$M = \dfrac{n}{V} = \dfrac{m}{M} \times \dfrac{{1000}}{V}$
Where m is mass of solute
M is the molar mass of solute
V is the volume of solution in mL
$\dfrac{5}{{342}} = \dfrac{{0.877}}{M}$
$\Rightarrow M = 0.877 \times \dfrac{{342}}{5}$
$\Rightarrow M = 59.98\;g\;mo{l^{ - 1}}$
Therefore the molecular weight of substance X is $59.98$

Hence the correct answer is option D.

Note: Here in the question cane sugar is a non-electrolyte which is non dissociated and this equation is valid. Also if there was an electrolyte was there in the process then we would include an i factor. It is also referred to as the van’t Hoff factor which measures the effect of solute on the given properties such as osmotic pressure.