A solution of mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ is 31.4\%$31.4\%$ mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ by mass and has a density of 1.25mathrmg / mL$1.25\mathrm{g / mL}$ . The molarity of the mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ solution is \_\_\_\_\_\_ mathrmM$\_\_\_\_\_\_ \mathrm{M}$ (nearest integer) [Given molar mass of mathrmH_2mathrmSO_4 = 98mathrmg mol^-1$\mathrm{H}_2\mathrm{SO}_4 = 98\mathrm{g mol}^{-1}$ ]

### Related Formula textMolarity (M) = frac\% text by mass times 10 times dM_w$$\text{Molarity } (M) = \frac{\% \text{ by mass} \times 10 \times d}{M_w}$$ ### Core Logic Let's assume we have 100 g of the solution. Mass of H_2SO_4$H_2SO_4$ in 100 g solution = 31.4text g$31.4\text{ g}$. Moles of H_2SO_4$H_2SO_4$ (n_textsolute$n_{\text{solute}}$) = frac31.498text mol$\frac{31.4}{98}\text{ mol}$. Volume of the solution (V$V$) can be found using density: V = fractextMass of solutiontextDensity = frac1001.25text mL$$V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100}{1.25}\text{ mL}$$ ### Step 1: Calculating Molarity Molarity is defined as moles of solute per liter of solution: M = fracn_textsoluteV(textin mL) times 1000$$M = \frac{n_{\text{solute}}}{V(\text{in mL})} \times 1000$$ M = frac31.4 / 98100 / 1.25 times 1000$$M = \frac{31.4 / 98}{100 / 1.25} \times 1000$$ M = frac31.4 times 1.2598 times 100 times 1000$$M = \frac{31.4 \times 1.25}{98 \times 100} \times 1000$$ M = frac39.2598 times 10$$M = \frac{39.25}{98} \times 10$$ M = 0.4005 times 10$$M = 0.4005 \times 10$$ M = 4.005text M$$M = 4.005\text{ M}$$ Rounding off to the nearest integer gives 4. ### Pattern Recognition Whenever percentage by mass (w/w$w/w$) and density (d$d$ in g/mL) are given, use the direct formula: M = frac\%(w/w) times d times 10M_w$M = \frac{\%(w/w) \times d \times 10}{M_w}$. This saves enormous time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry