To prepare 20 mL of 15 mM NaCl, how much NaCl would you need to weigh?
2. To prepare 5 mL of 1 mM glucose, how much glucose would you need to weigh? How do you prepare
1 mL of 1 mM NaCl using the NaCl solution you prepared in (1)?
3. How do you prepare 1 mL of 10 µM glucose using the glucose solution you prepared in (2)?
So repost I am not sure if I am doing any of these right
1M is not mole , its Molarity or amount of concentration .
please check on the definition once more .
To answer these questions, we need to understand the concepts of molarity and molecular weight. Molarity is a measurement of the concentration of a solute in a solution, expressed as the number of moles of the solute per liter of solution. Molecular weight refers to the mass of one mole of a substance and is expressed in grams.
1. To prepare 20 mL of 15 mM NaCl:
a) Start by determining the molecular weight of NaCl, which is 58.44 g/mol (sodium 22.99 g/mol + chlorine 35.45 g/mol).
b) To calculate the number of moles required, divide the desired concentration (15 mM) by 1000 to convert it to moles per milliliter.
15 mM NaCl = (15 mmol/L) x (20 mL/1000 mL) = 0.3 mmol NaCl
c) Finally, calculate the mass of NaCl needed by multiplying the number of moles by the molecular weight:
0.3 mmol NaCl x 58.44 g/mol = 17.53 mg NaCl
Therefore, you would need to weigh approximately 17.53 mg of NaCl.
2. To prepare 5 mL of 1 mM glucose:
a) Determine the molecular weight of glucose, which is 180.16 g/mol.
b) Convert the desired concentration to moles:
1 mM glucose = (1 mmol/L) x (5 mL/1000 mL) = 0.005 mmol glucose
c) Calculate the mass of glucose needed by multiplying the moles by the molecular weight:
0.005 mmol glucose x 180.16 g/mol = 0.9 mg glucose
Hence, you would need to weigh approximately 0.9 mg of glucose.
To prepare 1 mL of 1 mM NaCl using the NaCl solution prepared in (1):
a) We know that the concentration of the NaCl solution is 15 mM.
b) Convert the concentration to moles by multiplying by the volume:
15 mM NaCl x (20 mL/1000 mL) = 0.3 mmol NaCl
c) Since we want to prepare 1 mL, we can calculate the new concentration:
(0.3 mmol NaCl) / (20 mL) = 0.015 mmol/mL NaCl
d) Convert the new concentration back to millimolar:
0.015 mmol/mL NaCl x 1000 = 15 mM NaCl
Therefore, to prepare 1 mL of 1 mM NaCl, you would take 1 mL of the NaCl solution prepared in (1).
3. To prepare 1 mL of 10 µM glucose using the glucose solution prepared in (2):
a) We know that the concentration of the glucose solution is 1 mM.
b) Convert the concentration to moles by multiplying by the volume:
1 mM glucose x (5 mL/1000 mL) = 0.005 mmol glucose
c) Since we want to prepare 1 mL, we can calculate the new concentration:
(0.005 mmol glucose) / (5 mL) = 0.001 mmol/mL glucose
d) Convert the new concentration to micromolar:
0.001 mmol/mL glucose x 1000 = 1 µM glucose
Therefore, to prepare 1 mL of 10 µM glucose, you would dilute 1 mL of the glucose solution prepared in (2) to a total volume of 1 mL.
Go with the definition. 1M means 1 mol in 1 L of solution. So for #1, 15 mM is 0.015M so you want 0.015 mols in 1L. You only want 20 mL so you want
0.015 mols x (20 mL/1000 mL) = ? mols NaCl.
mols NaCl = grams NaCl/molar mass NaCl. You know mols and molar mass, solve for grams NaCl.
I get that. but what about how would you prepare 1mL of 1 mM NaCl using the NaCl solution you prepared in part 1?
You know the first one is 0.015. You want to dilute it to 0.01M. You want 1 mL of the 0.01 so
0.015 x mL = 0.01 x 1 and solve for mL of the 0.015. You may need a micropipet OR you can make up solutions and make serial dilutions.