Solving Chemistry Problem II: 0.005 M Carbonate Solution

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In summary, the conversation discusses how to calculate the mass of an unknown sample needed to prepare a 15 mL solution with a concentration of 0.005 M carbonate, assuming the weight percent of CaCO3 in the sample is 50%. The suggested approach is to first determine the number of moles of CaCO3 needed, using the formula moles = concentration x volume. Then, using the molar mass equation, the mass of CaCO3 can be calculated. Since the sample is assumed to be 50% CaCO3, the mass of the unknown sample can be found by doubling the calculated mass of CaCO3.
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higherme
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My question is:

Assuming the weight percent of CaCO3 is 50%, calculate the mass of unknown sample required to prepare a 15 mL solution that contains 0.005 M carbonate.

Well i know that in the sample of unknown, there half of them are CaCO3. I don't know what do to next...help please

Thank you
 
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  • #2
if you mean the mass percent of CaCO3 is 50% in an unknown sample then, try getting to how many moles of CaCO3 you would need to make a .005 M solution of Carbonate in 15mL
 
  • #3
that means i would need:
0.015 L x 0.005 mol/L = 7.5 E -5 moles of carbonate which means i neeed 7.5 E-5 moles of CaCO3

then... ?
 
  • #4
keep going, how do you get mass (grams) from moles? a great equation to know is MM (molar mass) = g/mol

then you have mass of CaCO3, if its 50% (or 1/2) of the mass (x) your unknown sample ...
 

1. How do I calculate the molarity of a 0.005 M carbonate solution?

To calculate the molarity of a solution, you need to know the moles of solute and the volume of the solution. In this case, the moles of carbonate can be calculated using the formula: moles = molarity x volume. So, for a 0.005 M carbonate solution with a volume of 1 liter, the moles of carbonate would be 0.005 moles.

2. How do I convert a concentration given in grams per liter to molarity?

To convert from grams per liter to molarity, you need to know the molar mass of the solute. First, calculate the moles of solute by dividing the mass in grams by the molar mass. Then, divide the moles by the volume in liters to get the molarity. For example, if you have 10 grams of carbonate (molar mass of 60 g/mol) in 1 liter of solution, the molarity would be 0.167 M (10 g / 60 g/mol / 1 L).

3. How do I dilute a 0.005 M carbonate solution to a lower concentration?

To dilute a solution, you need to add more solvent (usually water) to decrease the concentration. The new concentration can be calculated using the formula: M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. So, if you want to dilute a 100 mL solution of 0.005 M carbonate to 0.001 M, you would add 400 mL of water (0.005 M x 100 mL = 0.001 M x 500 mL).

4. What is the difference between molarity and molality?

Molarity (M) is a measure of the concentration of a solution in moles of solute per liter of solution. Molality (m) is a measure of the concentration of a solution in moles of solute per kilogram of solvent. Molarity takes into account the volume of the solution, while molality is independent of volume and is more accurate for calculations involving temperature changes.

5. How do I prepare a 0.005 M carbonate solution from a solid carbonate compound?

To prepare a solution from a solid compound, you need to calculate the grams of solute needed based on the desired molarity and volume. First, calculate the moles of solute using the formula: moles = molarity x volume. Then, use the molar mass of the compound to convert moles to grams. For example, to prepare 1 liter of 0.005 M carbonate solution from solid Na2CO3 (molar mass of 106 g/mol), you would need 0.53 grams of Na2CO3 (0.005 mol x 106 g/mol).

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