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What mass of KCl is required to prepare 383.00 mL of 0.500 M solution?
Wiki User
∙ 12y ago
I'll go straight to the equation,
to prepare 0.5 M of KCl in 383 ml, simply calculate it this way,
*Molar mass of KCl is 74.5513 g/mol
*0.5 M is also known as 0.5 mol/Litre
74.5513 g/mol x 0.5 mol/litre x 0.383 Litre = 14.2766 g
So, you need 14.2766 g of KCl to prepare 0.5 M in 383 ml solution.
Wiki User
∙ 12y ago
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