If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution,

1.	If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.
Answer» Mass of KOH = 5.6g No. of moles = \(\frac{5.6}{5.6}\) = 0.1 mol 1. Volume of the solution = 500 ml = 0.5 L 2. Volume of the solution = 1 L Molarity = (Number of moles of solute) / (Volume of soluation(in L) = \(\frac{0.1}{1}\) = 0.1 M 3. Volume of the solution = 1 L Molarity = M Molarity = \(\frac{0.1}{1}\) M

If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.

Answer»

Mass of KOH = 5.6g

No. of moles = \(\frac{5.6}{5.6}\) = 0.1 mol

1. Volume of the solution = 500 ml = 0.5 L

2. Volume of the solution = 1 L

Molarity = (Number of moles of solute) / (Volume of soluation(in L)

= \(\frac{0.1}{1}\) = 0.1 M

3. Volume of the solution = 1 L Molarity = M

Molarity = \(\frac{0.1}{1}\) M

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If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.

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