Question

How many grams of potassium acetate are present in a 54ml

solution of 2mEq/mL potassium acetate (M.W. =98 g/mol). Round to
the nearest hundredth. KC2H3O2 = K+1 + [C2H302]-1

Answer 1

The number of grams of potassium acetate present in the 54 ml solution of 2mEq/mL potassium acetate is approximately 10.58 grams.

Step-by-step explanation:

To calculate the number of grams of potassium acetate present in the solution, we need to use the equation:

[K+] = 2 * 0.054 L = 0.108 mol

The molar mass of potassium acetate (KC2H3O2) is 98 g/mol. Therefore, the number of grams of potassium acetate can be calculated as:

0.108 mol * 98 g/mol = 10.584 g

Round to the nearest hundredth, the number of grams of potassium acetate in the solution is 10.58 g.