Question

A potassium chloride solution that also contains 5% (m/V) dextrose is administered intravenously to treat some forms of malnutrition. The potassium ion concentration in this solution is 35 meq/L. Calculate the potassium ion concentration in units of mol/L.

Answer 1

Answer : The potassium ion concentration in units of mol/L is, $3.5\times 10^{-2}mol/L$

Explanation :

As we are given that the concentration of potassium ion in the solution is 35 meq/L. Now we have to determine the potassium ion concentration in units of mol/L.

First we have to convert meq/L to eq/L.

Conversion used :

1 meq/L = 10⁻³ eq/L

As, 1 meq/L concentration of potassium ion = 10⁻³ eq/L

So, 35 meq/L concentration of potassium ion = 35 × 10⁻³ eq/L

As we know that:

$\text{Moles}=\frac{\text{Gram equivalent}}{n_f}$

where,

$n_f$ = n- factor or valency of ion

$n_f$ for potassium ion = 1

So, the concentration of potassium ion = $\frac{35\times 10^{-3}mol/L}{1}=3.5\times 10^{-2}mol/L$

Therefore, the potassium ion concentration in units of mol/L is, $3.5\times 10^{-2}mol/L$

Answer 2

ci=oncentration of pottasium = 3.5 * 10^-2 mol/litre