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.

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 / 128.28 g mol^-1

                                                = 0.0055411 moles = 5.5411 x 10^-3 moles

Thus, 1 mL of nonane has 0.0055441 moles

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

Step : Preparation of final stock solution:

Since, 0.0055441 moles is present in 1 mL

So,      1 mole -           -           (1/ 0.0055441) ml
Or,        0.05 moles        -           (1/ 0.0055441) x 0.05 ml = 9.0185 mL

Take 9.0185 mL [0.718 g/ mL) x 9.0185 mL = 6.475 gram] sample in 1000 mL standard volumetric flask. Make the final upto the mark. It’s the desired solution.

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 = 5.9189 x 10^-3 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 1 L

Step: Preparation of final stock solution:

Since, 0.0059189 moles is present in 1 mL

So,      1 mole -           -           (1/ 0.0059189) ml
Or,        0.05 moles        -           (1/ 0.0059189) x 0.05 ml = 8.4475 mL

Take 8.4475 mL [0.84 g/ mL) x 8.4475 mL = 7.095 gram] sample in 1000 mL standard volumetric flask. Make the final upto the mark. It’s the desired solution.

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 = 5.4143 x 10^-3 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 1 L

Step : Preparation of final stock solution:

Since, 0.0054143 moles is present in 1 mL

So,      1 mole -           -           (1/ 0.0054143) ml
Or,        0.05 moles        -           (1/ 0.0054143) x 0.05 ml = 9.2378 mL

Take 9.2378 mL [0.861 g/ mL) x 9.2378 mL = 7.5911 gram] sample in 1000 mL standard volumetric flask. Make the final upto the mark. It’s the desired solution.