Question

The molarity of the solution when 4.9 g of H$_2$SO$_4$ is dissolved in 250 mL of solution is

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $0.4$

Hint:

Molarity Formula.

• $M = \fracnV$
• $n = \frac\textmass\textmolar mass$
• Convert mL to L by dividing by 1000.

Answer 1

The Correct Option is B

Given:

• Mass of H$_2$SO$_4$ = 4.9 g
• Volume of solution = 250 mL = 0.25 L
• Molar mass of H$_2$SO$_4$ = 98 g/mol

Moles of H$_2$SO$_4$: \[ \frac{4.9}{98} = 0.05~\text{mol} \] Molarity $M$: \[ M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.05}{0.25} = 0.2~\text{mol/L} \]

Answer 2

Step 1: Understanding the problem
We need to find the molarity of a solution prepared by dissolving 4.9 g of sulfuric acid (H₂SO₄) in 250 mL (0.250 L) of solution.

Step 2: Calculate the number of moles of H₂SO₄
Molar mass of H₂SO₄ = (2 × 1) + (32) + (4 × 16) = 2 + 32 + 64 = 98 g/mol
Number of moles = Mass / Molar mass = 4.9 g / 98 g/mol = 0.05 mol

Step 3: Calculate molarity
Molarity (M) = Number of moles of solute / Volume of solution in liters
M = 0.05 mol / 0.250 L = 0.2 M

Step 4: Conclusion
Therefore, the molarity of the solution is 0.2 M.