Chapter 3, Problem 3.23P

If 9 g of NaC1 were dissolved in 1 L of water, what would be concentration, in atoms per milliliter, of each of the constituent elements in the solution?

To determine

The concentration in atoms per millilitre of each of the constituent elements in the solution.

Answer to Problem 3.23P

Concentrations:

Number of atoms of Na added to the solution in 9 gm = $9.27\times {10}^{19}atomsNa/mL$

Number of atoms of Cl added to the solution in 9 gm = $9.27\times {10}^{19}atoms\text{ Cl}/mL$

Number of atoms of H in 1 millilitre of water = $6.69\times {10}^{22}atoms\text{ H}/mL$

Number of atoms of O in 1 millilitre of water = $3.34\times {10}^{22}atoms\text{ O}/mL$

Explanation of Solution

Given Information:

9 g of NaCl were dissolved in 1 L of water

Calculation:

Na = 22.99 g/mole

Cl = 35.45 g/mole

NaCl = 58.43 g/mole

Calculating the number of atoms of Na added to the solution in 9 gm,

  $(\frac{9g\text{ NaCl}}{1L})\times (\frac{\text{mol NaCl}}{58.43g\text{ NaCl}})\times (\frac{\text{1 mol Na}}{\text{1mol NaCl}})\times (\frac{6.02\times {10}^{23}\text{atoms Na}}{\text{1mol Na}})\times (\frac{1L}{1000mL}$ = $9.27\times {10}^{19}atomsNa/mL$

Calculating the number of atoms of Cl added to the solution in 9 gm,

  $(\frac{9g\text{ NaCl}}{1L})\times (\frac{\text{mol NaCl}}{58.43g\text{ NaCl}})\times (\frac{\text{1 mol Cl}}{\text{1mol NaCl}})\times (\frac{6.02\times {10}^{23}\text{atoms Cl}}{\text{1mol Cl}})\times (\frac{1L}{1000mL})$

  = $9.27\times {10}^{19}atoms\text{ Cl}/mL$

H = 1 g/mole

O = 16 g/mole

H₂O = 2(1 g/mole) + 16 g/mole = 18 g/mole

Calculating the number of atoms of H in 1 millilitre of water (assume 1 litter = 1000 g water),

  $(\frac{1000g{\text{ H}}_{2}O}{1L})\times (\frac{{\text{mol H}}_{2}O}{18g{\text{ H}}_{2}O})\times (\frac{\text{2 mol H}}{{\text{1mol H}}_{2}O})\times (\frac{6.02\times {10}^{23}\text{atoms H}}{\text{1mol H}})\times (\frac{1L}{1000mL})$ = $6.69\times {10}^{22}atoms\text{ H}/mL$

Calculating the number of atoms of O in 1 millilitre of water (assume 1 litter = 1000 g water),

  $(\frac{1000g{\text{ H}}_{2}O}{1L})\times (\frac{{\text{mol H}}_{2}O}{18g{\text{ H}}_{2}O})\times (\frac{\text{2 mol O}}{{\text{1mol H}}_{2}O})\times (\frac{6.02\times {10}^{23}\text{atoms O}}{\text{1mol O}})\times (\frac{1L}{1000mL})$ = $3.34\times {10}^{22}atoms\text{ O}/mL$

Conclusion:

The concentration, in atoms per millilitre, of each of the constituent elements in the solution is as follows.

Number of atoms of Na added to the solution in 9 gm = $9.27\times {10}^{19}atomsNa/mL$

Number of atoms of Cl added to the solution in 9 gm = $9.27\times {10}^{19}atoms\text{ Cl}/mL$

Number of atoms of H in 1 millilitre of water = $6.69\times {10}^{22}atoms\text{ H}/mL$

Number of atoms of O in 1 millilitre of water = $3.34\times {10}^{22}atoms\text{ O}/mL$