Sodium oxide reacts with water to produce sodium hydroxide. $20.0 \mathrm{~g}$ of sodium oxide is dissolved in $500 \mathrm{~mL}$ of water. Neglecting the change in volume, the concentration of the resulting $\mathrm{NaOH}$ solution is $\times 10^{-1}$ M. (Nearest integer)
$[$ Atomic mass : $\mathrm{Na}=23.0, \mathrm{O}=16.0, \mathrm{H}=1.0]$
$\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}$
$\frac{20}{62}$ moles
Moles of $\mathrm{NaOH}$ formed $=\frac{20}{62} \times 2$
$[\mathrm{NaOH}]=\frac{\frac{40}{62}}{\frac{500}{1000}}=1.29 \mathrm{M}=13 \times 10^{-1} \mathrm{M}$
(Nearest integer)
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