Question

Use the density and molecular weight of limonene, linalool, and nonane to calculate the volumes of the compounds you will need to make the stock solution of 50 mM. Also calculate the final concentration of those solutions after dilution.

Nonane: d = 0.718 g/mL, MW = 128.26, 99% pure

Limonene: d = 0.84 g/mL MW = 136.24, 96% pure

Linalool: d = 0.861 g/mL, MW = 154.25, 97% pure.

Preparing in 10mL volumetric flask.

Answer 1

Ans. 1. % purity = 99%.

Mass of 1 mL noname sample = density x volume = (0.718 g/ mL) x 1 mL = 0.718 g

Thus, actual amount of nonane in 0.718 g sample = 99% of 0.718 g = 0.71082 g

So, 1 ml of sample has 0.71082 g nonane.

Number of moles in 0.71082 g sample = mass/ molecular mass

                                                = 0.71082 g / 128.26 g mol^-1

                                                = 0.005542 moles

Thus, 1 mL of nonane has 0.005542 moles

Stock = 50 mM solution = 5.0 x 10^-2 M = 5.0 x 10^-2 moles in 1 L = 0.05 moles in 1000 mL

            Since, 1000 mL contains 0.05 moles

            So,        1 mL     -           (0.05 / 1000) moles

            So,        10 mL               (0.05 / 1000) x 10 moles = 0.0005 moles

Thus, 10 mL of 50 mM stock solution contains 0.0005 moles of the solute

# Preparation of final stock solution:

Since, 0.005542 mole is present in 1 mL

So,      1 mole -           -           (1/ 0.005542) ml
Or,        0.0005 moles    -           (1/ 0.005542) x 0.0005 ml = 0.0902197 mL

Mass of 0.0902197 mL nonane = density x volume

                                                = (0.718 g/ mL) x 0.0902197 mL = 0.06477 g

= 64.77 mg = 65 mg (approx.)

Take 65 mg sample in 10 mL standard volumetric flask. Make the final upto the mark. It’s 10 mL of 50 mM solution of nonane.

Re-check Concentration: Mass of solute = 0.06477 g

                                    Number of moles = 0.06477 g / 128.26 g mol^-1 = 0.00050505 moles

So, 10 mL has 0.00050505 moles solutes.

Moles of solute in 1000 mL = 0.0005050 moles x (1000 mL/ 10 mL) = 0.05 moles

So, 1000 mL (= 1 L) has 0.05 moles = 0.05 x 1000 millimoles = 50 millimoles      ;[1 mol = 10³ mmoles]

So, concentration = 50 mM in 1 L = 50 mM

Ans. 2. % purity = 96%.

Thus, actual amount of Limonene in 0.84 g sample = 96% of 0.84 g = 0.8064 g

So, 1 ml of sample has 0.8064 g Limonene.

Number of moles in 0.8064 g sample = mass/ molecular mass

                                                = 0.8064 g / 136.24 g mol^-1

                                                = 0.0059189 moles

Thus, 1 mL of limonene has 0.0059189 moles

Stock = 50 mM solution = 5.0 x 10^-2 M = 5.0 x 10^-2 moles in 1 L = 0.05 moles in 1000 mL

            Since, 1000 mL contains 0.05 moles

            So,        1 mL     -           (0.05 / 1000) moles

            So,        10 mL               (0.05 / 1000) x 10 moles = 0.0005 moles

Thus, 10 mL of 50 mM stock solution contains 0.0005 moles of the solute

# Preparation of final stock solution:

Since, 0.0059189 moles mole is present in 1 mL

So,      1 mole -           -           (1/ 0.0059189 moles) ml
Or,        0.0005 moles    -           (1/ 0.0059189 moles) x 0.0005 ml = 0.0844742 mL

Mass of 0.0844742 mL sample = density x volume

                                                = (0.84 g/ mL) x 0.0844742 mL = 0.070958 g

= 70.95 mg = 71 mg (approx.)

Take 71 mg sample in 10 mL standard volumetric flask. Make the final upto the mark. It’s 10 mL of 50 mM solution.

Re-check Concentration: Mass of solute = 0.070958 g

                                    Number of moles = 0.070958 g / 136.24 g mol^-1 = 0.0005208 moles

So, 10 mL has 0.0005208 moles solutes.

Moles of solute in 1000 mL = 0.0005208 moles x (1000 mL/ 10 mL) = 0.05 moles

So, 1000 mL (= 1 L) has 0.05 moles = 0.05 x 1000 millimoles = 50 millimoles

So, concentration = 50 mM in 1 L = 50 mM

Ans. 3. % purity = 97%.

Thus, actual amount of Linalool in 0.861 g sample = 97% of 0.861 g = 0.83517 g

So, 1 ml of sample has 0.83517 g Linalool.

Number of moles in 0.83517 g sample = mass/ molecular mass

                                                = 0.83517 g / 154.25 g mol^-1

                                                = 0.0054143 moles

Thus, 1 mL of Linalool has 0.0054143 moles

Stock = 50 mM solution = 5.0 x 10^-2 M = 5.0 x 10^-2 moles in 1 L = 0.05 moles in 1000 mL

            Since, 1000 mL contains 0.05 moles

            So,        1 mL     -           (0.05 / 1000) moles

            So,        10 mL               (0.05 / 1000) x 10 moles = 0.0005 moles

Thus, 10 mL of 50 mM stock solution contains 0.0005 moles of the solute

# Preparation of final stock solution:

Since, 0.0054143 moles mole is present in 1 mL

So,      1 mole -           -           (1/ 0.0054143 moles) ml
Or,        0.0005 moles    -           (1/ 0.0054143 moles) x 0.0005 ml = 0.092346 mL

Mass of 0.092346 mL sample = density x volume

                                                = (0.8461 g/ mL) x 0.092346 mL = 0.07951 g

= 79.51 mg = 80 mg (approx.)

Take 80 mg sample in 10 mL standard volumetric flask. Make the final upto the mark. It’s 10 mL of 50 mM solution.

Re-check Concentration: Mass of solute = 0.07951 g

                                    Number of moles = 0.07951 g / 154.25 g mol^-1 = 0.0005154 moles

So, 10 mL has 0.0005154 moles solutes.

Moles of solute in 1000 mL = 0.0005154 ­­moles x (1000 mL/ 10 mL) = 0.05 moles

So, 1000 mL (= 1 L) has 0.05 moles = 0.05 x 1000 millimoles = 50 millimoles

So, concentration = 50 mM in 1 L = 50 mM