Question

Concentrated aqueous sulfuric acid is 98% \[{{H}_{2}}S{{O}_{4}}\] by mass and has a density of 1.80\[gm{{L}^{-1}}\]. Volume of acid required to make one litre of 0.1M \[{{H}_{2}}S{{O}_{4}}\] is:
A. 5.55 mL
B. 11.10 mL
C. 16.65 mL
D. 22.20 mL

Answer 1

Hint: Calculate the volume of acid required by taking into account the data provided and use formula for normality to calculate the volume.

Complete step by step answer:
METHOD 1 –
We can directly use the data given in the question in the following way by relating normality and volume.
\[\text{Normality = }\dfrac{\text{weight }\!\!%\!\!\text{ x density x 10}}{\text{equivalent weight}}=\dfrac{\text{9}\text{.8 x 1}\text{.8 x 10}}{\text{49}}\text{ = 36N}\]
Since, \[{{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}\]
\[\Rightarrow \text{ 36 x }{{\text{V}}_{\text{1}}}\text{= 0}\text{.2 x 1000}\]
Therefore, \[{{\text{V}}_{\text{1}}}\text{ = }\dfrac{\text{0}\text{.2 x 1000}}{\text{36}}\text{ = 5}\text{.55 mL}\]

METHOD 2 –
According to the question, we have been given concentrated aqueous sulfuric acid which is 98% \[{{H}_{2}}S{{O}_{4}}\] by mass, which means –
\[\text{98 }\!\!%\!\!\text{ by weight }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}=\text{ }\dfrac{\text{98g }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}{\text{100g solution}}\]
We’ve also been given density of concentrated aqueous sulfuric acid; therefore, we can find volume of 100g solution by –
\[\text{Volume of 100g solution = }\dfrac{\text{mass}}{\text{density}}\text{=}\dfrac{\text{100}}{\text{1}\text{.8}}\text{=55}\text{.55mL}\]
\[\begin{align}
  & \text{For 0}\text{.1 M, we need 9}\text{.8g }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\
 & \therefore \text{ 55}\text{.55 mL acid = 98g} \\
 & \text{5}\text{.55 mL acid =9}\text{.8 g = 0}\text{.1 Molar acid} \\
\end{align}\]
Therefore, the answer is – option (a)

Note: Keep the following points in mind about Normality –
Symbol: N
Definition: “It is defined as the gram equivalent weight per litre of solution”.
Formula: \[\text{Normality = }\dfrac{\text{mass of solute (g)}}{\text{equivalent mass x volume of solution (L)}}\]
Unit: gram equivalent/ L