The mass of hydrogen (in grams) present in $1.0 \mathrm{~L}$ of pure water of density $1.0 \mathrm{~g}…
The mass of hydrogen (in grams) present in $1.0 \mathrm{~L}$ of pure water of density $1.0 \mathrm{~g} \mathrm{~cm}^{-3}$ is
$1.11 \times 10^2$
$5.55 \times 10^2$
$2.22 \times 10^2$
$3.33 \times 10^2$
Solution
Given, $V$ of water $=1 \mathrm{~L}=1000 \mathrm{~mL}$
Density of water $=1.0 \mathrm{~g} \mathrm{~cm}^{-3}$
To find, mass of hydrogen in water (in g)
$
\begin{array}{rlrl}
& & d & =\frac{m}{V} \text { (for water) } \\
\Rightarrow & & 1 & =\frac{m}{1000} \\
\therefore & m & =1000 \mathrm{~g}
\end{array}
$
Molecular mass of $\mathrm{H}_2 \mathrm{O}=(2 \times 1)+16=18$
$\because 18 \mathrm{~g}$ of water has $=2 \mathrm{~g}$ hydrogen
$\therefore 1000 \mathrm{~g}$ of water has $=\frac{2}{18} \times 1000 \mathrm{~g}$ of hydrogen
$
\begin{aligned}
& =111.11 \mathrm{~g} \text { of hydrogen } \\
& =1.11 \times 10^2 \mathrm{~g}
\end{aligned}
$