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if 45.1 mlml of 0.104 mm hclhcl solution is needed to neutralize a solution of kohkoh , how many grams of kohkoh must be present in the solutio

Sagot :

45.1 ml of 0.104 mm hcl solution is needed to neutralize a solution of koh , 0.294 grams of koh must be present in the solution

MHCl*VHCl = MKOH*VKOH = moles of KOH

moles of KOH = 0.116 M x (45.2/1000) L = 5.24 x 10^-3 mol

molar mass of KOH = 56.1 g/mol

mass of KOH = molar mass x moles = 56.1 x 5.24 x 10^-3 = 0.294 g

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