Question

A 5% solution of cane sugar (MW = 342) isisotonic with 1% solution of substance X. The molecular weight of X is

• A. 171.2
• B. 68.4
• C. 34.2
• D. 136.2

Answer 1

The Correct Option is B

To solve this problem, we need to use the concept of isotonic solutions and understand how to apply the formula for osmotic pressure in solutions. According to the problem, a 5% solution of cane sugar is isotonic with a 1% solution of substance X.

Isotonic solutions have the same osmotic pressure, which can be described by van't Hoff's equation for osmotic pressure:

\(\Pi = \frac{n}{V}RT\)

where \(\Pi\) is the osmotic pressure, \(n\) is the number of moles, \(V\) is the volume, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

For solutions of isotonicity, \(\frac{w_1}{M_1} \cdot \frac{1}{V_1}\) should be equal to \(\frac{w_2}{M_2} \cdot \frac{1}{V_2}\), where \(w_1\) and \(w_2\) are the weights of solutes, \(M_1\) and \(M_2\) are their molecular weights, and \(V_1\) and \(V_2\) are their volumes (which are the same here since percentages describe ratio in same volume). Therefore:

\(\frac{5}{342} = \frac{1}{M}\)

Here, \(5\%\) cane sugar solution has \(5\) grams of cane sugar in \(100\) mL of solution, and the molecular weight (MW) of cane sugar is \(342\).

Rearranging the equation to find \(M\), the molecular weight of substance X:

\(M = \frac{1 \times 342}{5} = 68.4\)

The molecular weight of substance X is 68.4. Thus, the correct answer is 68.4.