You have one liter of a 2M solution of compound x and a one liter solution of water how many milliliters of the 2M compound X solution and of water would you need to prepare the following dilute solutions of compound X?a) 100ml of a .2M compound X solution ml of 2M compound X solution + ml waterb) 10ml of a .2M compound X solution: ml of 2M compound X solution + ml water

The correct answers are explained below:

Explanation:

                                            M₁ V₁  = M₂ V₂-------(i)

where,  

M₁ is the molarity of Solution 1 (2 M Compound X solution),

M₂ is the molarity of Solution 2 (0.2 M solution of Compound X, that needs to be prepared from the 2 M Compound X solution).

V₁ is the volume of Solution 1 (Volume of the 2 M Compound X solution required),

V₂ is the volume of Solution 2 (Final Volume of the resulting 0.2 M solution of Compound X).

a) To prepare 100ml of a .2 M compound X solution 10 ml of 2 M compound X solution + 90 ml water.

Putting values in equation (i),

                             2 M V₁ = .2 M 100 millilitre

Therefore,            V₁ =  

So, the volume of 2M Compound X solution required is 10 millilitre, then the volume of water that needs to be added to make the final volume of the .2 M Compound X solution 100 millilitre is = 100 millilitre - 10 millilitre = 90 millilitre.

b) To prepare 10ml of a .2M compound X solution 1 ml of 2M compound X solution + 9 ml water.

Putting values in equation (i),

                             2 M V₁ = .2 M 10 millilitre

Therefore,            V₁ =  

So, the volume of 2 M Compound X solution required is 1 millilitre, then the volume of water that needs to be added to make the final volume of the .2 M Compound X solution 10 millilitre is = 10 millilitre - 1 millilitre = 9 millilitre.

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