Question

3. The same well water contains dissolved calcium (Ca), magnesium (Mg), sodium (Na) and potassium (K) at 5.0, 1.0, 136 and 1.2 mg/L, respectively. Determine these concentrations in units of mol/L.

Answer 1

Answer: The concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$, $4.20\times 10^{-5}mol/L$, $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively

Explanation:

To convert the mass from milligrams to grams, we use the conversion factor:

1 g = 1000 mg

To convert the given concentration fro grams to moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    .....(1)

• For Calcium:

Concentration given = 5.0 mg/L = $5.0\times 10^{-3}g/L$

We know that:

Molar mass of calcium element = 40 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{5.0\times 10^{-3}g/L}{40g/mol}\\\\\text{Concentration in mol/L}=1.25\times 10^{-4}mol/L$

• For Magnesium:

Concentration given = 1.0 mg/L = $1.0\times 10^{-3}g/L$

We know that:

Molar mass of magnesium element = 24 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.0\times 10^{-3}g/L}{24g/mol}\\\\\text{Concentration in mol/L}=4.20\times 10^{-5}mol/L$

• For Sodium:

Concentration given = 136 mg/L = $136\times 10^{-3}g/L$

We know that:

Molar mass of sodium element = 23 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{136\times 10^{-3}g/L}{23g/mol}\\\\\text{Concentration in mol/L}=5.91\times 10^{-3}mol/L$

• For Potassium:

Concentration given = 1.2 mg/L = $1.2\times 10^{-3}g/L$

We know that:

Molar mass of potassium element = 39 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.2\times 10^{-3}g/L}{39g/mol}\\\\\text{Concentration in mol/L}=3.08\times 10^{-5}mol/L$

Hence, the concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$, $4.20\times 10^{-5}mol/L$, $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively