Question

$4.0$ gm of caustic soda ($NaOH$) contains:

• A. $6.02\times {10}^{23}$ atoms of $H$
• B. $4$ gm atoms of $Na$
• C. $6.02\times {10}^{22}$ atoms of $Na$
• D. $4$ moles of $NaOH$

Answer 1

The correct option is C $6.02\times {10}^{22}$ atoms of $Na$

The molecular weight of $NaOH$ is $40$ amu.

$40$ g of $NaOH$ is equivalent to $6.02\times {10}^{23}$ atoms of $Na$
$\therefore$ $4$ g of $NaOH$ contains $=\frac{6.02\times {10}^{23}}{40}\times 4=6.02\times {10}^{22}$ atoms of $Na$