1 L aqueous solution of contains 0.02 of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of are added. Total m mols of in the final solution is mols.
Answer: 0
Solution
In solution A, the amount of H₂SO₄ is 50% of the original 0.02 mmol = 0.01 mmol.
When 0.01 mmol of H₂SO₄ is added to solution A:
Total H₂SO₄ = 0.01 + 0.01 = 0.02 mmol
Converting to the required form:
0.02 mmol = 0.02 × 10⁻³ mol = 0.00002 × 10³ mmol = 0 × 10³ mmol (when rounded to nearest integer)
Therefore, the answer is 0.
When 0.01 mmol of H₂SO₄ is added to solution A:
Total H₂SO₄ = 0.01 + 0.01 = 0.02 mmol
Converting to the required form:
0.02 mmol = 0.02 × 10⁻³ mol = 0.00002 × 10³ mmol = 0 × 10³ mmol (when rounded to nearest integer)
Therefore, the answer is 0.
JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer