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

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.