A 5% solution of cane sugar with a molecular weight of 342 is isotonic with 1% solution of a substance X. What is the molecular weight of X?

Cane sugar:

W1 = 5 g

V1 = 100 ml = 0.1 l

M1 = 342

&nbps;

X:

W2 = 1 g

V2 = 100 ml = 0.1 l

M2 = ?

We know that, for isotonic solution, C1 = C2

W1/M1V1 = W2/M2V2

5/(341×0.1) = 1/(M2x0.1)

M2 = 342/5 = 68.4

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