Question

Find the molarity of water. Given: $\rho = 1000\, kg /m^{3}$
[Report your answer after rounding off to the nearest integer].

Answer 1

Density of water $=1000\, Kg / m ^{3}=1 \,g / mL$
Volume of water $=1\, L =1000 \,mL$
Mass of $1.0\, L$ water $=1000\, g$
Molarity $=\frac{\text { mass of water }}{\text { molar mass of water }} \times \frac{1000}{ v _{ mL }}$
$\Rightarrow$ Molarity $=\frac{1000}{18} \times \frac{1000}{1000}=55.5 \,mol \,L ^{-1}$
Since before $5$ there is an odd digit so while rounding off we will add $1$ to the next digit.
So, final answer is $56 .$