A3.682 g sample of kclo3 is dissolved in enough water to give 375. ml of solution. what is the chlorate ion concentration in this solution?

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

ugh fiel iente of insistí is did kiwis jaj

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole