Question

1L aqueous solution of $H_2SO_4$ contains 0.02 m mol $H_2SO_4$. 50% of this solution is diluted with deionized water to give 1L solution (A). In solution (A), 0.01 m mol of $H_2SO_4$ are added. Total m mols of $H_2SO_4$ in the final solution is ______ $× 10^3$ m mols.

Answer 1

Initially one litre contains 0.02 mole.
∴ 50% of this solution will contains 0.01 m mol.
After adding 0.01 mol, final solution will contain 0.02 m mol of $H_2SO_4$.
= 0.02 m mol
= 0.00002 x 103 m molCorrect answer should be 0.