Question

The mass of hydrogen in grams present in \(1.0\ \text{L}\) of pure water of density \(1.0\ \text{g cm}^{-3}\) is:

• A. \(1.11\times10^2\)
• B. \(5.55\times10^2\)
• C. \(2.22\times10^2\)
• D. \(3.33\times10^2\)

Hint:

In water, the mass fraction of hydrogen is: \[ \frac{2}{18}=\frac{1}{9} \] So, mass of hydrogen can be found by multiplying the mass of water by \(\frac{1}{9}\).

Answer 1

The Correct Option is A

Step 1: Calculate the mass of water.
Given, \[ \text{Volume}=1.0\ \text{L} \] Since, \[ 1\ \text{L}=1000\ \text{cm}^3 \] and density of water is \[ 1.0\ \text{g cm}^{-3} \] Mass of water: \[ \text{Mass}=\text{Density}\times\text{Volume} \] \[ =1.0\times1000 \] \[ =1000\ \text{g} \]

Step 2: Find the fraction of hydrogen in water.
Molecular formula of water is \[ H_2O \] Molar mass of water: \[ =2(1)+16 \] \[ =18\ \text{g mol}^{-1} \] Mass of hydrogen in \(18\ \text{g}\) of water: \[ =2\ \text{g} \] Therefore, fraction of hydrogen in water: \[ =\frac{2}{18} \] \[ =\frac{1}{9} \]

Step 3: Calculate mass of hydrogen in \(1000\ \text{g}\) water.
\[ \text{Mass of hydrogen} = 1000\times\frac{1}{9} \] \[ =111.11\ \text{g} \] \[ =1.11\times10^2\ \text{g} \]

Step 4: Final conclusion.
Therefore, the mass of hydrogen present is \[ \boxed{1.11\times10^2\ \text{g}} \]