Question

How many atoms of hydrogen are present in 2.92 g of water?

Answer 1

$\boxed{1.954\times 10^{23}\text{ atoms}}$of hydrogen are present in 2.92 g of ${{\text{H}}_{\text{2}}}{\text{O}}$ .

Further Explanation:

In ${{\text{H}}_2}{\text{O}}$ , both hydrogen and oxygen are non-metals. Oxygen has six valence electrons whereas hydrogen has only one valence electron. Both share two electrons with each other and two covalent bonds are formed between oxygen and two hydrogen atoms. This bond is formed due to sharing of the electrons and therefore ${{\text{H}}_2}{\text{O}}$ is a covalent compound.

${{\text{H}}_2}{\text{O}}$ is a covalent compound. So 1 mole of ${{\text{H}}_2}{\text{O}}$ contains 1 mole of oxygen and 2 moles of hydrogen.  

The formula to calculate the moles of water is s follows:  

${\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}}=\frac{{{\text{Given mass of }}{{\text{H}}_{\text{2}}}{\text{O}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{2}}}{\text{O}}}}$ .......(1)

The given mass of water is 2.92 g.  

The molar mass of water is 18 g/mol.  

Substitute these values in equation (1)

$\begin{aligned}{\text{Moles of }}{{\text{H}}_2}{\text{O}}&= \left({2.92\;{\text{g}}} \right)\left({\frac{{1\;{\text{mol}}}}{{18\;{\text{g}}}}}\right)\\&={\mathbf{0}}{\mathbf{.1622}}\;{\mathbf{mol}}\\\end{aligned}$

The number of molecules in 1 mol of ${{\text{H}}_2}{\text{O}}$ is $6.022 \times {10^{23}}$ .

So the number of molecules in 0.1622 mol of ${{\text{H}}_2}{\text{O}}$ is calculated as follows:

$\begin{aligned}{\text{Number of water molecules}} &= \left( {0.1622\;{\text{mol}}} \right)\left({\frac{{6.022 \times {{10}^{23}}\;{\text{molecules}}}}{{1\;{\text{mol}}}}} \right)\\&={\mathbf{0}}{\mathbf{.9768 \times 1}}{{\mathbf{0}}^{{\mathbf{23}}}}\;{\mathbf{molecules}}\\\end{aligned}$

Each molecule of ${{\text{H}}_2}{\text{O}}$ has 2 atoms of hydrogen.

So the number of hydrogen atoms in $0.9768 \times {10^{23}}\;{\text{molecules}}$ of water is calculated as follows;

$\begin{aligned}{\text{Number of atoms of hydrogen}}&=\left({\frac{{2\;{\text{atoms}}}}{{1\;{\text{molecule}}}}}\right)\times\left( {0.9768 \times {{10}^{23}}{\text{molecules}}}\right)\\&=1.954\times {10^{23}}\;{\text{atoms}}\\\end{aligned}$

$\mathbf{1.954}\mathbf{\times}\mathbf{10^{23}}\textbf{ atoms}$ of hydrogen are present in 2.92 g of ${{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}$.

Learn more:  

1. Identification of ionic bonding: brainly.com/question/1603987  

2. Which is a covalent compound? brainly.com/question/2083444  

Answer details:

Grade: Senior School  

Subject: Chemistry  

Chapter: Mole concept  

Keywords: H2O, covalent compound, molar mass, atoms, hydrogen, 18 g/mol, valence electrons, 2.92 g, moles, given mass.  

Answer 2

The molecular weight of water is 18.01528 g/mol.
So in 2.92 grams there are 2.92/18.01528 = 0.1621 mol of particles.

1 mol contains 6,02214 × 10^23 particles by definition.

So the nr of H2O molecules is 0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.

Every molecule has 2 H atoms, so you have to double that.

2* 0,9761 × 10^23 = 1.952 × 10^23.