Question

How many grams of potassium iodine, kl,must be added to 500.0 grams of water to produce a 6.00% solution ?

Answer 1

grams KI needed = 31.9 grams

Step-by-step explanation:

? g KI + 500 g water => 6.0% KI solution

let x = grams KI needed.

Percent = part/total x 100% = (x/x+500)·100% = 6.0%

Solve for 'x' ...

x / x + 500 = 6/100 = 0.06 => x = 0.06(x + 500) => x = 0.06x + 30

x - 0.06x = 30 => 0.94x = 30 => x = 30/0.94 = 31.9 gms KI needed.