Question

A 3.682 g sample of kclo3 is dissolved in enough water to give 375. ml of solution. what is the chlorate ion concentration in this solution?

Answer 1

mKClO₃: 39g+35,5g+(16g×3) = 122,5 g/mol

1 mol -- 122,5g
X mol -- 3,682g
X = 3,682/122,5
X = 0,03 mol

375 ml = 0,375 l = 0,375 dm³

C = n/V
C = 0,03/0,375
C = 0,08 mol/dm³

[ClO₃-] = [KClO₃] = 0,08 mol/dm³

Answer 2

Answer: The chlorate ion concentration in the solution is 0.080 M

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

We are given:

Mass of solute (potassium chlorate) = 3.682 g

Molar mass of potassium chlorate = 122.55 g/mol

Volume of solution = 375 mL

Putting values in above equation, we get:

`\text{Molarity of solution}=(3.682g* 1000)/(122.55g/mol* 375mL)\\\\\text{Molarity of solution}=0.080M`

The chemical equation for the ionization of potassium chlorate follows:

`KClO_3\rightarrow K^++ClO_3^-`

1 mole of potassium chlorate produces 1 mole of potassium ions and 1 mole of chlorate ions

So, concentration of chlorate ions in the solution = 0.080 M

Hence, the chlorate ion concentration in the solution is 0.080 M