Question

100 ml of 10V $H_2{O}_2$ solution is heated. The evolved gas is completely reacted with Ca to form CaO. The aqueous solution of Cao is neutralised by 50 ml of $H_{2}S{O}_4$ solution. The molarity of $H_{2}SO4$ is:

• A. 2.43 M
• B. 1.78 M
• C. 1.55 M
• D. 2.78 M

Answer 1

The correct option is B 1.78 M

Given,

$100ml$ of $10V$ $H_2{O}_2$ solution

That means $10V$ $H_2{O}_2$ is

$1L$ of $H_2{O}_2$ produces $10L$ $O_2$

$\Rightarrow 100ml$ of $H_2{O}_2$ produces $100\times {10}^{-3}\times 10L$ of $O_2$

$\Rightarrow 100ml$ $10V$ $H_2{O}_2$ produces $1L$ of $O_2⟶\left(1\right)$

Again this $O_2$ is reacted with $Ca$

The reaction involved is

$C{a}_2+O_2⟶2CaO$

The reaction shows $1$ mole $C{O}_2$ & $1$ mole $O_2$ produces $2$ mole $CaO$

From (1) & $O_2$ reacted is calculated by

$1$ mole $⟶22.4L$ of $O_2$

$x$ mole$⟶1L$ of $O_2$

$x=\frac{1}{22.4}$ moles of $O_2$

$\Rightarrow \frac{1}{22.4}$ moles of $O_2$ reacts with $\frac{1}{22.4}$ moles $C{a}_2$ to produce $\frac{2}{22.4}$ moles $CaO$

Again, this $Cao$ is reacted with $H_{2}S{O}_4$

The reaction is

$CaO+H_{2}S{O}_4⟶CaS{O}_4+H_{2}O$

$\Rightarrow 1$ mole $CaO$ consumes $1$ mole $H_{2}S{O}_4$

$\Rightarrow \frac{2}{22.4}$ mole $CaO$ consumes $\frac{2}{22.4}$ mole $H_{2}S{O}_4$.

$\frac{2}{22.4}$ mole $H_{2}S{O}_4$

Volume of $H_{2}S{O}_4=50ml$

$\therefore$ Molarity of $H_{2}S{O}_4=No.ofmoles\times \frac{1000}{Vinml}$

$=\frac{2}{22.4}\times \frac{1000}{50}$

$=1.78M$.