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 nM. 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%.

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   ------- sample 1

Stock = 50 nM solution = 5.0 x 10^-8 M = 5.0 x 10^-8 moles in 1 L.

Now, 1000 mL of stock has 50 nmoles solute       --- sample 2

            1 mL     -           - (50 nmoles / 1000 mL) = 5.0 x 10^-4 nmoles = 5.0 x 10^-13 moles –

Preparing necessary dilution: Since, the concentration of the stock solution is very low, we need to make two dilutions to make its preparation practically feasible.

Step 1: Take 1 mL of original sample in 1L standard volumetric flask and make the volume upto mark with distilled water. This solution is 1:1000 diluted. Thus, concentration of this solution (say, solution A) = 5.5411 x 10^-6 moles in 1 L = 5.5411 x 10^-6 M  

Step 2: Preparation of final stock solution:

M1V1 = M2V2 --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 (solution A)

            M2= molarity of solution 2, V2= volume of solution 2 (solution B, Stock, to be prepared)

Suppose, we need to prepare 100 mL of stock solution

                5.5411 x 10^-6 M x V1 = 5.0 x 10^-8 x 100 mL              

                Or, V1 = (5.0 x 10^-8 M x 100 mL) / 5.5411 x 10^-6 M = 0.902

Thus, take 902 uL (microliter) of solution A in 100-mL standard volumetric flask and make the volume upto the mark with distilled water. The resultant solution in 50 nM.

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   ------- sample 1

Stock = 50 nM solution = 5.0 x 10^-8 M = 5.0 x 10^-8 moles in 1 L.

Preparing necessary dilution: Since, the concentration of the stock solution is very low, we need to make two dilutions to make its preparation practically feasible.

Step 1: Take 1 mL of original sample in 1L standard volumetric flask and make the volume upto mark with distilled water. This solution is 1:1000 diluted. Thus, concentration of this solution (say, solution A) = 5.9189 x 10^-6 moles in 1 L = 5.9189 x 10^-6 M  

Step 2: Preparation of final stock solution:

M1V1 = M2V2 --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 (solution A)

            M2= molarity of solution 2, V2= volume of solution 2 (solution B, Stock, to be prepared)

Suppose, we need to prepare 100 mL of stock solution

                5.9189 x 10^-6 M x V1 = 5.0 x 10^-8 x 100 mL              

                Or, V1 = (5.0 x 10^-8 M x 100 mL) / 5.9189 x 10^-6 M = 0.844

Thus, take 844 uL (microliter) of solution A in 100-mL standard volumetric flask and make the volume upto the mark with distilled water. The resultant solution in 50 nM.

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 / 136.24 g mol^-1

                                                = 0.0054143 moles = 5.4143 x 10^-3 moles   ------- sample 1

Stock = 50 nM solution = 5.0 x 10^-8 M = 5.0 x 10^-8 moles in 1 L.

Preparing necessary dilution: Since, the concentration of the stock solution is very low, we need to make two dilutions to make its preparation practically feasible.

Step 1: Take 1 mL of original sample in 1L standard volumetric flask and make the volume upto mark with distilled water. This solution is 1:1000 diluted. Thus, concentration of this solution (say, solution A) = 5.4143 x 10^-6 moles in 1 L = 5.4143 x 10^-6 M  

Step 2: Preparation of final stock solution:

M1V1 = M2V2 --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 (solution A)

            M2= molarity of solution 2, V2= volume of solution 2 (solution B, Stock, to be prepared)

Suppose, we need to prepare 100 mL of stock solution

                5.4143 x 10^-6 M x V1 = 5.0 x 10^-8 x 100 mL              

                Or, V1 = (5.0 x 10^-8 M x 100 mL) / 5.4143 x 10^-6 M = 0.923

Thus, take 923 uL (microliter) of solution A in 100-mL standard volumetric flask and make the volume upto the mark with distilled water. The resultant solution in 50 nM.