Question

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

• 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. $6\times {10}^{23},1.5\times {10}^{23},3\times {10}^{28}$
• D. $1.5\times {10}^{23},1.5\times {10}^{23},6\times {10}^{23}$

Answer 1

Answer: (A)

$1.5\times {10}^{23},0.75\times {10}^{23},3\times {10}^{23}$

10 g of $NaOH$ (molar mass 40 g/mol) $=\frac{10g}{40g/mol}=0.25moles$
$2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$
0.25 moles of $NaOH$ will react with $\frac{0.25}{2}=0.125$ moles of $H_{2}S{O}_4$.
Number of molecules of $H_{2}S{O}_4=0.125\times 6.023\times {10}^{23}=7.53\times {10}^{22}$
Number of atoms of $H=2\times 7.53\times {10}^{22}=1.5\times {10}^{23}$
Number of atoms of $S=1\times 7.53\times {10}^{22}=0.75\times {10}^{23}$
Number of atoms of $O=4\times 7.53\times {10}^{22}=3\times {10}^{23}$