Question

Mass of $70$% $H_{2}S{O}_4$ solution (by mass) required for neutralisation of $1$ mol of $NaOH$ is:

• A. 35 g
• B. 70 g
• C. 90 g
• D. 50 g

Answer 1

Answer: (B)

The explanation for the correct option:

Option (B) 70 g

• The neutralization reaction between $H_{2}S{O}_4$ and $NaOH$ gives sodium sulfate and water as a product.

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

• According to the reaction, $49$ g of $H_{2}S{O}_4$ is required for one mole of NaOH.
• The equation for Mass% is given as:$Mass\%=\frac{Givenmass}{molarmass}\times 100$
• Here, Mass % is given as 70%, and half a mole of $H_{2}S{O}_4$ is reacting with $NaOH$.

• So, $70$% mass of $H_{2}S{O}_4$ have weight of $\frac{49\times 100}{70}=70gm$

Hence, Option(B) 70 gm is the mass of 70% $H_{2}S{O}_4$ required for the neutralization of 1 mole of NaOH