To what volume should you dilute 133 mL of an 7.90 M copper (II) chloride solution so that 51.5 mL of the diluted solution contains 4.49 grams of copper (II) chloride? Report your answer to 3 sigfigs, not including units when you type in your answer

See explanation below

Explanation:

In order to do this, we first need to know the molecular mass of the Copper(II) chloride to get the moles.

The Copper(II) chloride formula is CuCl₂, and the atomic mass of each element is:

Cu = 63.55 g/mol;   Cl = 35.45 g/mol

The molecular mass will be:

MM = 63.55 + (35.45*2) = 134.45 g/mol

let's see how many moles we have in the 7.90 M solution. This is, to know how many grams we have in that solution and stablish a relation between this mass and the desired mass:

moles = 7.90 * 0.133 = 1.0507 moles

mass = 1.0507 * 134.45 = 141.27 g

This means that we have 141.27 g of copper chloride initially, and we need to get 4.49 g of the solution. When a solution is dilluted, means that the solute remains constant and the amount of solvent is increased. Therefore we just need to figure out the volume of the initial solution that can contain the desired mass. This is done by this:

If 133 mL contains 141.27 g, then with 4.49 g in how much volume is contained?

V = 4.49 * 133/141.27 = 4.23 mL

This means that if we take 4.23 mL of the initial solution and dillute it to 51.5 mL, this solution will contain the 4.49 g of copper chloride.

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