Question

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

• A. 49 gm
• B. 98 gm
• C. 70 gm
• D. 34.3 gm

Answer 1

Answer: (C)

70 gm

Answer 2

H2SO4 + 2NaOH $\longrightarrow$ Na2SO4 + 2H2O

for 1 mole NaOH $\frac { 1 } { 2 }$ mole H2SO4 required

70 g H2SO4 in 100 g solution

g H2SO4 in = $\frac { 100 } { 70 }$ × 49 = 70 g solution.