To what volume should you dilute 60.0 mL of a 4.50 M KI solution so that 30.0 mL of the diluted solution contains 3.25 g of KI?

You should dilute to 414.1 mL

Explanation:

Let's apply the dilution formula to solve this:

M conc . Vol conc = M dil . Vol dil

Molarity for concentrated solution → 4.50 M

Volume for concentrated solution → 60 mL

Molarity for diluted solution → Unknown

Volume for diluted solution → Answer

Let's determine the molarity for diluted solution with data given.

We convert the mass of KI to moles

3.25 g . 1 mol/166 g = 0.0195 moles

Let's convert the volume from mL to L

30 mL . 1 L/1000mL = 0.030 L

Molarity (mol/L) = 0.0195 mol /0.030L = 0.652 M

We replace the data in the formula

4.50 M . 60 mL = 0.652 M . Volume for diluted solution

(4.50M . 60mL) / 0.652M = Volume for diluted solution

414.1 mL = Volume for diluted solution

413mL

Explanation:

Molar Mass of KI = 39 + 127 = 166g/mol

Mass conc of KI = 3.25g

Number of mole = Mass conc/Molar Mass

Number of mole = 3.25/166 = 0.0196 mol

But recall

Molarity = number of mole / Volume

Number of mole = 0.0196 mol

Volume = 30mL = 30/1000 = 0.03L

Molarity = 0.0196/0.03

Molarity = 0.653M

C1 = 4.5M

V1 = 60mL = 60/1000 = 0.06L

C2 = 0.653M

V2 =?

C1V1 = C2V2

4.5 x 0.06 = 0.653 x V2

V2 = (4.5 x 0.06) / 0.653

V2 = 0.413L

Converting to mL

0.413 x 1000 = 413mL

it is stable as it is in group 0 which have full outer shells of electrons

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.