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A $5\% $ solution of cane sugar is isotonic with $0.877\% $ of $X$. The molecular weight of substance $X$ is:
$A.59.98$
$B.119.96$
$C.95.58$
$D.126.98$

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Last updated date: 13th Jun 2024
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Answer
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Hint:The concentration of the given solution can be calculated equating the mass of the given solute inside the solution and the molecular weight of the solute. As the isotonic solution has the same concentration as there is no net movement of the molecules in the solution due to the presence of no gradient.

Complete step by step answer:
The \[5\% \] solution of cane sugar is taken here to be mixed, which means $5g$ of cane sugar is mixed in $100ml$ of the solution. The given substance $X$ has the mixture of \[5\% \] solutions of cane sugar isotonic with the $0.877\% $ of the substance $X$.
Therefore, the isotonicity of the solutions means that the concentrations are equal and hence it can be proved by the formula: ${C_1} = {C_2}$. Here ${C_1}$ is the concentration of the cane sugar solution whole ${C_2}$ is the concentration of the substance $X$. The concentration of a given solution can be calculated using: $C = \dfrac{m}{{MW \times 100}}$ where the $m$ is the mass of the solute and $MW$ is the molecular weight. In the denominator $100$ is taken as the percentage value of the mass is given here for the solution. The molecular weight of sugar is the same as that of glucose which is \[342g/mol\]. The molecular weight of substance $X$ needs to be calculated here and hence for the equation, it can be taken as $X$. Therefore, if both the solutions have equal concentrations then for the equation can be arranged according to ${C_1} = {C_2}$ and using this
$\dfrac{5}{{342 \times 100}} = \dfrac{{0.877}}{{X \times 100}}$
 $ \Rightarrow \dfrac{5}{{34200}} = {\kern 1pt}\dfrac{{0.877}}{{100X}}$
$ \Rightarrow 500X = 0.877 \times 34200$
$ \Rightarrow 500X = 29993.4$
 $ \Rightarrow X = \dfrac{{29993.4}}{{500}}$
 $ \Rightarrow X = 59.98$
The given molecular weight of the solution $X$ as calculated from the equation of isotonicity is $59.98g/mol$.

Hence, option A is correct.

Note:
The concentration is the same for the two solutions under the given condition of isotonicity. The isotonic solution means no flow of molecules only which is why the conditions can be equated with one another.