How many grams of solid KCl are needed to prepare 250.0 mL of 0.235 M solution?
As, molarity = mass×1000/molar mass×volume(ml)
Molar mass of KCl = 74.55 g/mol
So, 0.235 = m×1000/74.55×250
m = 0.235×74.55×250/1000
= 4.38 g
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