Final answer:
The molecular mass of a non-volatile solute X, which is isotonic with a 10% urea solution, is calculated using the proportionality between mass and moles of substances in isotonic solutions. By equalizing the moles of urea and X and knowing the mass of X is 6%, the molecular mass of X is determined to be 36 g/mol. Therefore, the correct option is C.
Step-by-step explanation:
The question asks for the molecular mass of a non-volatile solute X, given that a 10% urea solution is isotonic with a 6% solution of X. Isotonic solutions have the same osmotic pressure, which implies they have equal numbers of solute particles per unit volume. Knowing that urea (NH2)2CO has a molar mass of 60 g/mol, we can set up a proportion to find the molar mass of X, using the fact that the concentrations given are weight/volume percentages.
Considering a 100 mL solution for both solutes for simplicity, the mass of urea would be 10 g (10% of 100 mL), which corresponds to 10 g / 60 g/mol = 0.1667 mol. For the solute X in a similar 100 mL solution, the mass would be 6 g (6% of 100 mL). Since the solutions are isotonic, the number of moles of X must be the same as that of urea, 0.1667 mol. Thus, the molar mass of X can be calculated: Molar mass of X = Mass of X / Moles of X = 6 g / 0.1667 mol, which equals approximately 36 g/mol. Option C, 36 g/mol, is the correct answer.