Answer
5.29 M
Work Step by Step
1. To find the molarity, we would require the number of moles in 1 liter of $H_{2}$$SO_{4}$ .
2. Density = Specific Gravity $\times$ 1 g/mL
3. Therefore, the density of the $H_{2}$$SO_{4}$ solution = 1.305 g/mL
4. So we will have 1305 grams of $H_{2}$$SO_{4}$ in 1 L solution as 1.305 g/mL $\times $ 1000 mL = 1305 g
5. Now, 39.77% $H_{2}$$SO_{4}$ means 39.77 grams of pure $H_{2}$$SO_{4}$ in every 100 grams of solution.
6. This means we have 1305 g $\times$ $\frac{39.77 g}{100 g}$ = 519 g $H_{2}$$SO_{4}$
7. Dividing the available mass of $H_{2}$$SO_{4}$ with its molar mass = $\frac{519 g}{98 g/mol}$ = 5.29 mol
8. Since we have 5.29 mol $H_{2}$$SO_{4}$ in 1 L of solution, we have 5.29 M $H_{2}$$SO_{4}$ solution.
