Question

142 g of sodium sulphate $\left(N{a}_{2}S{O}_4\right)$ has:

• A. 2 mole of $N{a}_{2}S{O}_4$
• B. $12.044\times {10}^{23}$ sodium ions
• C. 1 mole of $N{a}_{2}S{O}_4$
• D. $6.022\times {10}^{23}$ sulphur atoms

Answer 1

Answer: (B), (C), (D)

The correct options are
B $12.044\times {10}^{23}$ sodium ions
C 1 mole of $N{a}_{2}S{O}_4$
D $6.022\times {10}^{23}$ sulphur atoms

$142g$ of ${Na}_2{SO}_4=(23\times 2)+32+(16\times 4)$

$=46+32+64=142g$

$\therefore$ moles of ${Na}_2{SO}_4=\frac{142}{142}=1$ mole of ${Na}_2{SO}_4$

$1$ moles of ${Na}_2{SO}_4$ contains $1$ mole of $S$ atoms and $2$ moles of $Na$ atoms

$\therefore$ it contains $1\times 6.022\times {10}^{23}$ atoms of $S$

and $2\times 6.022\times {10}^{23}$ atoms of $Na$

$\Rightarrow 12.044\times {10}^{23}$ atoms of $S$