Two elements A and B form compounds having molecular formula and . When dissolved in 20 g of , 1 g o — Solutions and Colligative Properties Chemistry Question
Question
Two elements A and B form compounds having molecular formula $\text{AB}_2$ and $\text{AB}_4$. When dissolved in 20 g of $\text{C}_6\text{H}_6$, 1 g of $\text{AB}_2$ lowers the freezing point by 2.55 K, whereas 1.0 g of $\text{AB}_4$ lowers it by 1.7 K. The molar depression constant for benzene is $5.1 \text{ K-kg mol}^{-1}$. The atomic masses of A and B are
💡 Solution & Explanation
Molar mass $M = (K_f \times w \times 1000) / (\Delta T_f \times W)$. For $\text{AB}_2$, $M_1 = (5.1 \times 1 \times 1000) / (2.55 \times 20) = 5100 / 51 = 100 \text{ g/mol}$. For $\text{AB}_4$, $M_2 = (5.1 \times 1 \times 1000) / (1.7 \times 20) = 5100 / 34 = 150 \text{ g/mol}$. Let A and B be atomic masses. $A + 2B = 100$ and $A + 4B = 150$. Subtracting yields $2B = 50 \Rightarrow B = 25$. Substituting back, $A + 50 = 100 \Rightarrow A = 50$. Therefore, correct answer is A.