hi, i was wondering if anyone could help me with this problem? i am having some difficulty with it. thanks.
a).how many grams of Al(NO3)3 are needed to prepare 300mL of a 1.071 M solution?
b).if you further dilute this 300mL to a new final volume of 5.0L, calculate the final molarity.
molarity = #mols/Liter of solution.
grams = #mols x molar mass.
So use equation 1, plug in M and L and calculate # mols. Then go to eqution 2 and plug in #mols and molar mass and calculate grams.
ok, i got this for a)68.376 g and b) 0.064 Molars. is it correct?
I calculated 68.43 g for (a) which round to 68.4 as does your answer. For b, I obtained 0.06426 which rounds to 0.0643 but the 5.0 L has only two significant figures; therefore, this would round to 0.064 M.
Of course, I can help you with that! To determine the number of grams of Al(NO3)3 needed to prepare a 1.071 M solution in 300 mL and calculate the final molarity when diluted to a new final volume of 5.0L, we need to follow a step-by-step process.
Step 1: Determine the molar mass of Al(NO3)3.
Al(NO3)3 consists of one aluminum atom (Al) and three nitrate ions (NO3). The molar masses of the elements are:
Aluminum (Al): 26.98 g/mol
Nitrogen (N): 14.01 g/mol
Oxygen (O): 16.00 g/mol
Using the periodic table, we can calculate the molar mass of Al(NO3)3 as follows:
Al(NO3)3 = 1(Al) + 3(NO3)
= 1(26.98 g/mol) + 3(14.01 g/mol + 3(16.00 g/mol)
= 1(26.98 g/mol) + 3(62.03 g/mol)
= 26.98 g/mol + 186.09 g/mol
= 213.07 g/mol
So, the molar mass of Al(NO3)3 is 213.07 g/mol.
Step 2: Calculate the number of moles of Al(NO3)3 needed.
We can use the formula: concentration (M) = moles/volume (L) to calculate the number of moles.
Moles = concentration × volume
Moles = 1.071 mol/L × 0.3 L
Moles = 0.3213 mol
Step 3: Calculate the mass of Al(NO3)3 needed.
Mass = moles × molar mass
Mass = 0.3213 mol × 213.07 g/mol
Mass = 68.46 g
Therefore, you would need 68.46 grams of Al(NO3)3 to prepare a 1.071 M solution in 300 mL.
Now, let's proceed to the second part of your question:
Step 4: Calculate the concentration (molarity) of the diluted solution.
We can use the formula: moles = concentration × volume to calculate the number of moles of the solute in the diluted solution. Since the solution is diluted to a new final volume of 5.0 L:
Moles = concentration × volume
Moles = 1.071 mol/L × 0.3 L = 0.3213 mol
Step 5: Calculate the final molarity of the diluted solution.
We can use the formula: concentration (M) = moles/volume (L) to calculate the final molarity:
Molarity = moles/volume
Molarity = 0.3213 mol/5.0 L
Molarity = 0.06426 M
Therefore, the final molarity of the solution, when diluted to a new final volume of 5.0 L, is 0.06426 M.