The mass of 70 % H2SO4 required for neutralisation of 1 mole of NaOH is

Question

The mass of 70 % H2SO4 required for neutralisation of 1 mole of NaOH is (A) 49 g (B) 98 g (C) 70 g (D) 34 g

Tardigrade Admin · Accepted Answer

H2SO4 furnishes two protons, so 1/2 mole of H2SO4 is required for complete neutralization of 1 mole of NaOH. So, for 100 % H2SO4, 49 g of H2SO4 is required ∴ For 70 % H2SO4 (49× 70/100)=34 g is required

Q. The mass of $70%$ $H_{2} S O_{4}$ required for neutralisation of $1$ mole of $N a O H$ is

$49 g$

$98 g$

$70 g$

$34 g$

$H_{2} S O_{4}$ furnishes two protons, so 1/2 mole of $H_{2} S O_{4}$ is required for complete neutralization of 1 mole of NaOH.
So, for $100% H_{2} S O_{4}, 49 g$ of $H_{2} S O_{4}$ is required
$∴$ For $70% H_{2} S O_{4} \frac{49 \times 70}{100} = 34 g$ is required

Q. The mass of 70% H2​SO4​ required for neutralisation of 1 mole of NaOH is

49g

98g

70g

34g