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Question: A \(5\% \) solution of cane sugar is isotonic with \(0.877\% \) of \(X\). The molecular weight of su...

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

Explanation

Solution

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%5\% solution of cane sugar is taken here to be mixed, which means 5g5g of cane sugar is mixed in 100ml100ml of the solution. The given substance XX has the mixture of 5%5\% solutions of cane sugar isotonic with the 0.877%0.877\% of the substance XX.
Therefore, the isotonicity of the solutions means that the concentrations are equal and hence it can be proved by the formula: C1=C2{C_1} = {C_2}. Here C1{C_1} is the concentration of the cane sugar solution whole C2{C_2} is the concentration of the substance XX. The concentration of a given solution can be calculated using: C=mMW×100C = \dfrac{m}{{MW \times 100}} where the mm is the mass of the solute and MWMW is the molecular weight. In the denominator 100100 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/mol342g/mol. The molecular weight of substance XX needs to be calculated here and hence for the equation, it can be taken as XX. Therefore, if both the solutions have equal concentrations then for the equation can be arranged according to C1=C2{C_1} = {C_2} and using this
5342×100=0.877X×100\dfrac{5}{{342 \times 100}} = \dfrac{{0.877}}{{X \times 100}}
534200=0.877100X\Rightarrow \dfrac{5}{{34200}} = {\kern 1pt}\dfrac{{0.877}}{{100X}}
500X=0.877×34200\Rightarrow 500X = 0.877 \times 34200
500X=29993.4\Rightarrow 500X = 29993.4
X=29993.4500\Rightarrow X = \dfrac{{29993.4}}{{500}}
X=59.98\Rightarrow X = 59.98
The given molecular weight of the solution XX as calculated from the equation of isotonicity is 59.98g/mol59.98g/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.