How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?

ANSWERS

2015-11-03 23:37:21

C = n/V n = C×V n = 4,41M × 1,25L n = 5,5125 mol mKI: 39+127 = 166 g/mol 1 mol --------- 166g 5,5125 mol --- X X = 166×5,5125 = 915,075g KI :)

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