Question

10g of $NaOH$ required certain amount of $H_{2}S{O}_4$ for complete neutralisation. Calculate the number of atoms of hydrogen, sulphur and oxygen in $H_{2}S{O}_4$ used.

• A. $1.5\times {10}^{23},0.75\times {10}^{23},3\times {10}^{23}$
• B. $0.75\times {10}^{23},0.75\times {10}^{23},6\times {10}^{23}$
• C. $6\times {10}^{23},1.5\times {10}^{23},3\times {10}^{23}$
• D. $1.5\times {10}^{23},1.5\times {10}^{23},6\times {10}^{23}$

Answer 1

The correct option is A $1.5\times {10}^{23},0.75\times {10}^{23},3\times {10}^{23}$

The balanced equation for the reaction can be written as:

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

10g of $NaOH=$ $\frac{10}{40}$ = 0.25 moles

2 moles of $NaOH$ requires = 1 mole of $H_{2}S{O}_4$

0.25 moles of $NaOH$ requires = $\frac{1}{2}\times 0.25$ mol = 0.125 moles of $H_{2}S{O}_4$.

Each molecule of $H_{2}S{O}_4$ contain 2 H atoms, 1 S atom and 4 O atoms.

Now,

0.125 moles of $H_{2}S{O}_4$ contains = $2\times 6.02\times {10}^{23}\times 0.125$ = $1.5\times {10}^{23}$ atoms of hydrogen.

0.125 moles of $H_{2}S{O}_4$ contains = $1\times 6.02\times {10}^{23}\times 0.125$ = $0.75\times {10}^{23}$ atoms of sulphur

0.125 moles of $H_{2}S{O}_4$ contains = $4\times 6.02\times {10}^{23}\times 0.125$ = $3.01\times {10}^{23}$ atoms of oxygen

Therefore, $\left(A\right)$ is the correct answer.