The mass of 70 % H2SO4 required for neutralisation of 1 mol of NaOH.

Question

The mass of 70 % H2SO4 required for neutralisation of 1 mol of NaOH. (A) 49 g (B) 98 g (C) 70 g (D) 34.3 g

Tardigrade Admin · Accepted Answer

H2SO4+2NaOH → Na2 SO4+2H2O for 1 mole NaOH (1/2) mole H2SO4 required Molar mass of H2SO4 = 98 g 70 g H2SO4 in 100 g solution (98/2) g H2SO4 in =(100/70) × 49=70 g solution.

Q. The mass of $70% H_{2} S O_{4}$ required for neutralisation of $1$ mol of NaOH.

$49$ g

$98$ g

$70$ g

$34.3$ g

$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$
for $1$ mole $N a O H \frac{1}{2}$ mole $H_{2} S O_{4}$ required
Molar mass of $H_{2} S O_{4} = 98$ g
$70$ g $H_{2} S O_{4}$ in $100$ g solution
$\frac{98}{2} g H_{2} S O_{4} in = \frac{100}{70} \times 49 = 70$ g solution.

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

49 g

98 g

70 g

34.3 g