how many grams of Kl are needed to prepare 25.0ml of a 0.750M solution
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Answer:
2.988g
Explanation:
no.of mol of solution = (25 ml)×(0.750M)/1000
= 0.018
no.of mol of KI = 0.018 = given mass/ molarmass
= given mass/166
given mass = 166×0.018g
= 2.988g
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