Question

The density of sodium borohydride is $1.074{\text{g/ cm}}^3.3.91\text{gm}$ of sodium borohydride contains $2.50\times {10}^{23}$ atoms of $H.$ The number of moles of $H$-atoms present in $28.0{\text{cm}}^3$ of sodium borohydride is:

• A. $3.192$
• B. $2.03$
• C. $1.67$
• D. $1.92$

Answer 1

The correct option is A $3.192$

Weight of sodium borohydride in $28.0{\text{cm}}^3$
$=28\times 1.074=30.072\text{gm}$
$\because 3.91\text{gm of sodium borohydride has moles of}$
$H$- atoms = $\frac{2.50\times {10}^{23}}{6.023\times {10}^{23}}$
Moles of $H$- atoms in $30.072$ $gm$ of sodium borohydride = $\frac{2.50\times {10}^{23}}{6.023\times {10}^{23}}\times \frac{30.072}{3.91}$
= 3.192 mol