When 1~mathrmg$1~\mathrm{g}$ each of compounds AB and mathrmAB_2$\mathrm{AB}_2$ are dissolved in 15~mathrmg$15~\mathrm{g}$ of water separately, they increased the boiling point of water by 2.7~mathrmK$2.7~\mathrm{K}$ and 1.5~mathrmK$1.5~\mathrm{K}$ respectively. The atomic mass of A (in amu) is times 10^-1$\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is 0.5~mathrmK~kg~mol^-1$0.5~\mathrm{K~kg~mol^{-1}}$)

### Related Formula Delta T_mathrmb = K_mathrmb cdot m = K_mathrmb cdot left( fracw_textsoluteM_textsolute cdot frac1000w_textsolvent right)$$\Delta T_{\mathrm{b}} = K_{\mathrm{b}} \cdot m = K_{\mathrm{b}} \cdot \left( \frac{w_{\text{solute}}}{M_{\text{solute}}} \cdot \frac{1000}{w_{\text{solvent}}} \right)$$ ### Core Logic Both dissolved compounds are non-electrolytes (van 't Hoff factor i = 1$i = 1$). We calculate the molar masses of mathrmAB$\mathrm{AB}$ and mathrmAB_2$\mathrm{AB_2}$ individually, then solve for the individual atomic masses of elements A and B. ### Step 1: Determine Molar Mass of AB Given Delta T_mathrmb = 2.7~mathrmK$\Delta T_{\mathrm{b}} = 2.7~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5~mathrmK~kg~mol^-1$K_{\mathrm{b}} = 0.5~\mathrm{K~kg~mol^{-1}}$: 2.7 = 0.5 times frac1M_mathrmAB times frac100015$$2.7 = 0.5 \times \frac{1}{M_{\mathrm{AB}}} \times \frac{1000}{15}$$ M_mathrmAB = frac0.5 times 100015 times 2.7 = frac50040.5 approx 12.3457~mathrmg~mol^-1$$M_{\mathrm{AB}} = \frac{0.5 \times 1000}{15 \times 2.7} = \frac{500}{40.5} \approx 12.3457~\mathrm{g~mol^{-1}}$$ ### Step 2: Determine Molar Mass of AB_2 Given Delta T_mathrmb = 1.5~mathrmK$\Delta T_{\mathrm{b}} = 1.5~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5$K_{\mathrm{b}} = 0.5$: 1.5 = 0.5 times frac1M_mathrmAB_2 times frac100015$$1.5 = 0.5 \times \frac{1}{M_{\mathrm{AB_2}}} \times \frac{1000}{15}$$ M_mathrmAB_2 = frac0.5 times 100015 times 1.5 = frac50022.5 approx 22.2222~mathrmg~mol^-1$$M_{\mathrm{AB_2}} = \frac{0.5 \times 1000}{15 \times 1.5} = \frac{500}{22.5} \approx 22.2222~\mathrm{g~mol^{-1}}$$ ### Step 3: Solve for Atomic Mass of A Let the atomic masses of elements A and B be a$a$ and b$b$ respectively: a + b = 12.3457 quad text--- (1)$$a + b = 12.3457 \quad \text{--- (1)}$$ a + 2b = 22.2222 quad text--- (2)$$a + 2b = 22.2222 \quad \text{--- (2)}$$ Subtracting equation (1) from (2): b = 22.2222 - 12.3457 = 9.8765~mathrmamu$$b = 22.2222 - 12.3457 = 9.8765~\mathrm{amu}$$ Substituting b$b$ back into equation (1): a = 12.3457 - 9.8765 = 2.4692~mathrmamu$$a = 12.3457 - 9.8765 = 2.4692~\mathrm{amu}$$ Expressing a$a$ in the requested format (times 10^-1$\times 10^{-1}$): a = 24.692 times 10^-1 approx 25 times 10^-1$$a = 24.692 \times 10^{-1} \approx 25 \times 10^{-1}$$ ### Pattern Recognition Mathematical consistency checks: Since mathrmAB_2$\mathrm{AB_2}$ has more atoms than mathrmAB$\mathrm{AB}$ of identical constituent mass, its molar mass should be higher, leading to a smaller elevation of boiling point for the same mass dissolved, which perfectly matches our 2.7~mathrmK rightarrow 1.5~mathrmK$2.7~\mathrm{K} \rightarrow 1.5~\mathrm{K}$ change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions