Question

10. A 5% of cane sugar is isotonic with 0.877% solution of substance X. Find the molecular weight of X. [Ans. 60 g/mol]​

Answer 1

Answer:

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Explanation:

Let M g/mol be the molecular weight of substance X. The molecular weight of cane sugar is 342 g/mol.

Isotonic solutions have same osmotic pressure.

The osmotic pressure is given by the expression, π = CRT.

$π_{cane \: sugar }=π _X \\ C_{ cane \: sugar }RT=C_ XRT \\ C_{cane \: sugar } =C _X$

Here, concentration is in moles per litre. But we are given the concentration in weight / volume percent.

$C_{cane \: sugar }=C _ X \\ \\ ( \frac{5}{342}) _{cane \: sugar } =(\frac{0.877}{M})_X \\ \\ M=0.877 \times \frac{5}{342}$

M=59.98 g/mol.