Question

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

Answer 1

Answer:

The correct answers are explained below:

Explanation:

• Molarity of a solution can be defined as the concentration of solute in a given solution. It is expressed as the number of moles (calculated as the ratio between the amount to the solute added to the solvent to the molecular weight of the solute) of a given solute that is dissolved in one litre of the solvent.
• The given solution's Concentration  = 2 M.
• Volume of the given solution = 1 litre = 1000 millilitre.
• According to the equation of Molarity,

                                            M₁ $\times$ V₁  = M₂ $\times$ 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 $\times$ V₁ = .2 M $\times$ 100 millilitre

Therefore,            V₁ =  $\frac{.2\times100}{2} = 10 millilitre.$

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 $\times$ V₁ = .2 M $\times$ 10 millilitre

Therefore,            V₁ =  $\frac{.2\times10}{2} = 1 millilitre.$

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.