Question

Question

If 5.6 g of KOH is present in (a) 500 mL and (b) 1 litre of solution, calculate them molarity of each of these solutions.


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Answer 1

Answer:

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Answer 2

0.2  M  and 0.1 M is the molarity of 500 ml  and  1 litre solution.

Given:

KOH= 5.6 g

500 mL and 1 litre.

To find:

Molarity of each of the solution.

Solution:

The molar mass of KOH = 39.1 + 16 + 1

                                        = 56.1 g/mol

(a) For 500 mL solution:

As we know

Number of moles = mass / molar mass

So,

Number of moles of KOH =5.6 g / 56.1 g/mol

                                           = 0.1 mol

Volume in litre= 500 mL

                        = 0.5 L

So ,

Molarity (M) = 0.1 mol / 0.5 L

                    = 0.2 M

So the Molarity of 500 ml solution is 0.2 M.

(b) Molarity of 1 litre solution:

We know,

Number of moles of KOH = 0.1 mol

Given,

Volume = 1 L

So with this data we can find moarity of this solution ,

Now,

Molarity (M) = 0.1 mol / 1 L

                   = 0.1 M

So 0.1 M is the molarity of 1 litre solution .

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