Question

The molar mass of a solute ${\text{X}}$ in ${\text{gm mo}}{{\text{l}}^{ - 1}}$ , if it's $1\%$ solution is isotonic with {\text{5% }} solution of cane sugar${\text{(molar mass}} = 342{\text{gm mo}}{{\text{l}}^{ - 1}})$ , is
A.$68.4$
B.$34.2$
C.$136.2$
D.$171.2$

Answer 1

Isotonic solutions are those solutions which have the same concentration or same osmotic pressure. We need to first calculate the number of moles using mass percentage and then we will calculate the concentration in molarity and equate both the concentration to get the molar mass.
Formula used: ${\text{Molarity}} = \dfrac{{{\text{no}}{\text{. of moles }}}}{{{\text{volume of solution}}({\text{ml}})}} \times 1000$
${\text{no}}{\text{. of moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
Percentage by mass is defined as the mass of solute that is present in 100 gram of solution.
Mass percentage of ${\text{X}}$ is given to us that is $1\%$ , so 1 g of ${\text{X}}$ is contained in 100 g of the solution.
Mass percentage of cane sugar is {\text{5% }} , so 5 g of cane sugar is contained in 100 g of the solution.
${\text{moles of X}} = \dfrac{{{\text{mass of X}}}}{{{\text{molar mass of X}}}}$
Mass of ${\text{X}}$ is 1 g and the molar mass of ${\text{X}}$ we have to calculate let this be M.
Putting the values we will get the number of moles of ${\text{X}}$ as:
${\text{moles of X}} = \dfrac{1}{{\text{M}}}$
Similarly, Mass of cane sugar is 5 g and Molar mass of cane sugar is given that is $342{\text{gm mo}}{{\text{l}}^{ - 1}}$.
Putting the values we will get the number of moles of cane sugar as:
${\text{moles of cane sugar}} = \dfrac{5}{{342}}$
Now density of water is $1{\text{ g m}}{{\text{l}}^{ - 1}}$, hence mass of water will be same as volume of water which is same in both the cases that is 100.
Hence using the formula of molarity and equating them we will get:
$\dfrac{1}{{\text{M}}} \times \dfrac{{1000}}{{100{\text{ mL}}}} = \dfrac{5}{{342}} \times \dfrac{{1000}}{{100{\text{ mL}}}}$
Rearranging we will get
${\text{M}} = \dfrac{{342}}{5} = 68.5{\text{ g mo}}{{\text{l}}^{ - 1}}$

The correct option is A.

Note:
The concentration is defined in terms of percentage also. There are two ways of representing the concentration in terms of percentage that is percentage by mass and percentage by volume. When either of them is not specified in the question we will consider that percentage by mass is given to us.