Question

A 5% solution of cane sugar (Mol wt. = 342) is isotonic with 1% solution of substance X. The molecular weight of X is:
A. 171.2
B. 68.4
C. 34.2
D. 136.2

Answer 1

Hint: We know; Isotonic solution is a solution when its osmotic concentration is the same as that of another solution. So, here concentration of cane sugar will be equal to the concentration of the substance. With the help of concentration molecular weight could be known.

Complete step by step answer:
1. Now, as we know about the isotonic solutions. Accordingly, osmotic pressure of cane sugar is equal to that of substance X, i.e. $\pi$ $_1$= $\pi$ $_2$
Thus, C$_1$T$_1$ = C$_2$T$_2$, it represents the osmotic concentration.
At the constant temperature C$_1$ = C$_2$.

2. Now, molar concentration of cane sugar = $\dfrac{W_1}{M_1 \times V_1}$ , W$_1$ is 5%, given Mol wt. (M$_1$)= 342, V$_1$ is 1% ; So, $\dfrac{5 \times 1000}{342 \times 100}$ = $\dfrac{50}{342}$

3. Further, molar concentration of X =$\dfrac{W_2}{M_2 \times V_2}$, here let the Mol wt. (M$_2$) of X be $m$; So, $\dfrac{1\times 1000}{m \times 100}$ = $\dfrac{10}{m}$
Accordingly, $\dfrac{50}{342}$ = $\dfrac{10}{m}$, then m = 68.4.

S o, the molecular weight of X is 68.4; the correct option is B.

Additional information:
Tonicity is a term related to the osmotic pressure gradient, it is the water potential of two solutions separated by a semipermeable membrane. It is influenced only by the solutes that cannot cross the membrane.

Note: Don’t get confused between the steps followed. Just remember at the constant temperature molar concentration of one substance is equal to that of another substance. Then apply the formula of concentration. The molecular weight can be found, even the weight is given in percentage, converting it to the numbers.