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lawyer [7]
3 years ago
9

Dilution question In many of the experiments, you will be asked to prepare a standard solution by diluting a stock solution. You

will be specifically asked how you will do the dilution steps during the review before you start work. Here is a homework question to get you to think about this process. You have a 1000 ug/mL stock solution of lead (Pb2+). Describe how you would prepare a 100 ppb standard solution from this stock solution. Rules: No single dilution step can be greater than 1:50 You have 25.00 mL volumetric flasks and 10.00 mL volumetric flasks. You have access to three pipet dispensers. One dispenser delivers between 1 and 5 mL, the second dispenser delivers between 100 and 1000 , and the third dispenser delivers between 20 and 200 HL
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

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tangare [24]

Answer:

(a) cesium bromide (CsBr): 9.15 grams

(b) calcium sulfate (CaSO4):  5.85 grams

(c) sodium phosphate (Na3PO4): 7.05 grams

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Explanation:

<u>(a) cesium bromide (CsBr):</u>

Molar mass of CsBR = 212.81 g/mol

Number of moles = molarity * volume

Number of moles = 0.100 M *0.43 L

Number of moles = 0.043 moles

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<u>(b) calcium sulfate (CaSO4):</u>

Molar mass of CaSO4 = 136.14 g/mol

Mass of CaSO4 required = moles * Molar mass

Mass of CaSO4 required = 0.043 moles * 136.14 g/mol

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<u>(c) sodium phosphate (Na3PO4):</u>

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 required = moles * Molar mass

Mass of Na3PO4 required = 0.043 moles * 163.94 g/mol

Mass of Na3PO4 required = 7.05 grams

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Molar mass of Li2Cr2O7 = 229.87 g/mol

Mass of Li2Cr2O7 required = moles * Molar mass

Mass of Li2Cr2O7 required = 0.043 moles * 229.87 g/mol

Mass of Li2Cr2O7 required = 9.88 grams

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Molar mass of K2C2O4 = 166.22 g/mol

Mass of K2C2O4 required = moles * Molar mass

Mass of K2C2O4 required = 0.043 moles * 166.22 g/mol

Mass of K2C2O4 required = 7.15 grams

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