Help me set them up and then check my answers?
Round your answer to the correct number of significant digits and show work.
1. How many grams of NaI would be used to produce a 2.0 M solution with a volume of 10.00 L?
2. What is the molarity of a 2.50 x 102 cm3 solution containing 450.0 g of C6H12O6 ?
3. How many moles of Sr(NO3)2 would be used in the preparation of 20.50 L of a 30.5 M solution?
1. How many mols do you want? That's mols = M x L = ?
Then mols = g/molar mas. Substitute mols and molar mass and solve for grams. Then round to the right number of s.f.
Here is a site that will help you do that.
http://www.chemteam.info/SigFigs/SigFigs.html
2. mols = grams/molar mass
Then M = mols/L solution. Round to correct number of s.f.
3. The reverse of #2.
Post your work if you get stuck.
Sure! I can help you set up the calculations and then check your answers.
1. To calculate the number of grams of NaI needed to produce a 2.0 M solution with a volume of 10.00 L, you can use the formula:
grams = moles * molar mass
First, let's calculate the moles of NaI needed:
moles = Molarity * volume (in liters)
moles = 2.0 M * 10.00 L
moles = 20.0 moles
Next, we need to find the molar mass of NaI, which is the sum of the atomic masses of sodium (Na) and iodine (I):
molar mass (NaI) = atomic mass (Na) + atomic mass (I)
molar mass (NaI) = 22.99 g/mol + 126.9 g/mol
molar mass (NaI) = 149.89 g/mol
Now, we can calculate the grams of NaI needed:
grams = moles * molar mass
grams = 20.0 moles * 149.89 g/mol
grams = 2997.8 g
Therefore, you would need 2997.8 grams of NaI to produce a 2.0 M solution with a volume of 10.00 L.
Now, let's move on to the next question.
2. To calculate the molarity of a solution containing 450.0 g of C6H12O6 in a volume of 2.50 x 10^2 cm3, you need to convert the volume from cm3 to liters.
2.50 x 10^2 cm3 = 2.50 x 10^-1 L (since 1 cm3 = 10^-3 L)
Now, let's calculate the moles of C6H12O6:
moles = mass / molar mass
molar mass (C6H12O6) = (6 * atomic mass (C)) + (12 * atomic mass (H)) + (6 * atomic mass (O))
molar mass (C6H12O6) = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
molar mass (C6H12O6) = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
molar mass (C6H12O6) = 180.18 g/mol
moles = 450.0 g / 180.18 g/mol
moles = 2.498 moles (rounded to 3 significant digits)
Now, let's calculate the molarity:
Molarity = moles / volume (in liters)
Molarity = 2.498 moles / 2.50 x 10^-1 L
Molarity = 9.992 M (rounded to 4 significant digits)
Therefore, the molarity of the solution is 9.992 M.
Now, let's move on to the last question.
3. To calculate the moles of Sr(NO3)2 needed to prepare a 30.5 M solution with a volume of 20.50 L, you can use the formula:
moles = Molarity * volume (in liters)
moles = 30.5 M * 20.50 L
moles = 625.25 moles
Therefore, you would need 625.25 moles of Sr(NO3)2 to prepare a 30.5 M solution with a volume of 20.50 L.
I hope this helps! Let me know if you have any other questions.