Question

A 3.682 g sample of KClO3 is dissolved in enough water to give 375.0 mL of solution. What is the chlorate ion concentration in this solution?

Answer 1

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

Explanation:

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

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:

The chemical equation for the ionization of potassium chlorate follows:

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

Answer 2

Answer:

The concentration of the chlorate ion in the solution is $0.08 M$.

Explanation:

Given,

The mass of potassium chlorate, $KClO_{3}$, m = $3.682g$

The volume of the solution, i.e., water, V = $375ml$ = $0.375 L$

The concentration of the chlorate ion =?

As we know,

• Potassium chlorate dissolves into water to form potassium ions ($K^{+}$) and chlorate ions ($ClO_{3}^{-}$).
• $KClO_{3} (s)\rightarrow K^{+} (aq) + ClO_{3}^{-}(aq)$

From the reaction, we can say that one mole of $KClO_{3}$ form one mole of $ClO_{3}^{-}$ ions.

Therefore,

• The number of moles of $KClO_{3}$ = The number of moles of $ClO_{3}^{-}$ ions.

Now, firstly, we have to calculate the number of moles from the equation given below:

• $n = \frac{m}{M}$

Here,

• n = The number of moles
• m = Mass
• M = Molar mass of $KClO_{3}$ = $122.55g/mol$

After putting the given values in the equation, we get:

• $n = \frac{3.682}{122.55}$ = $0.03 mol$

Now, we have to calculate the molarity of chlorate ion by the equation given below:

• Molarity = $\frac{n}{V}$

Here,

• n = The number of moles of the solute, i.e., chlorate ions
• V = The volume of the solution

After putting the values in the equation, we get:

• Molarity = $\frac{0.03}{0.375}$ = $0.08M$

Hence, the concentration of chlorate ions in the solution = $0.08M$.