How many grams of kcl are present in 85.0 ml of 2.10 m kcl? how many grams of kcl are present in 85.0 ml of 2.10 m kcl? 3.02 133 13.3 6.34 none of the above?

Molarity (M) = moles of the solute (mol) / Volume of the solution (L)

The molarity of the given KCl solution = 2.10 M
Volume of the given KCl solution = 85.0 mL

Hence,
         2.10 M = Moles of KCl / 85.0 x 10⁻³ L
Moles of KCl = 2.10 M x 85.0 x 10⁻³ L
                     = 0.1785 mol

Moles (mol) = Mass (g) / Molar mass (g mol⁻¹)

Molar mass of KCl = 74.56 g mol⁻¹

Hence, 
  0.1785 mol = Mass of KCl / 74.56 g mol⁻¹
Mass of KCl = 0.1785 mol x 74.56 g mol⁻¹
                    = 13.30 g

Hence, mass of given KCl in 85.0 mL is 13.30 g.

Mgo2 is the oxidation state of manganese oxide