Question

1.00 mL of a 2.95 X10^-4 M solution of oleic acid is diluted wth 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid are 5.00 mL of solution B? (Molar mass for oleic acid=282g/mo)

Answer 1

Answer – Given, 1.00 mL of a 2.95 X10^-4 M solution of oleic acid

Volume of petroleum ether = 9.00 mL for solution A.

For solution –B – volume of solution = 2.00 mL and volume of petroleum ether = 8.00 mL

We need to calculate the grams of oleic acid are 5.00 mL of solution B

First we need to calculate the moles of oleic acid

We know, moles = molarity * volume (L)

Moles of oleic acid = 2.95 X10^-4 M * 0.001 L

                                = 2.95 X10^-7 moles

We need to calculate the moalrity of oleic acid in diluted solution

M1 = 2.95 X10^-4 M, V1= 1.00 mL, M2 = ? , V2 = 1.0+9.00 = 10.0 mL

M2 = M11V1 / V2

       = 2.95 X10^-4 M * 1.00 mL / 10.00 mL

       = 2.95 X10^-5 M

So solution A has concentration of oleic acid is 2.95 X10^-5 M

Now concentration of oleic acid in solution B –

M1 = 2.95 X10^-5 M, V1 = 2.00 mL , V2 = 10 mL , M2 = ?

M2 = M11V1 / V2

       = 2.95 X10^-5 M * 2.00 mL / 10.00 mL

       = 5.90 X10^-6 M

So moles of 5.00 mL of solution B = 0.005 L * 5.90 X10^-6 M

                                                         = 2.95*10^-8 moles

So, mass of oleic acid = 2.95*10^-8 moles * 282 g/mol

                                    = 8.32*10^-6 g

So, 8.32*10^-6 g of oleic acid are 5.00 mL of solution B.