1.)How many grams would you need of a sample known to be 99.81% AgN03 by mass?
You are asked to prepare 150.0 mL of 3.15×10−2 M AgNO3.
2.) An aqueous solution is 6.40% methanol CH3OH by mass, with d = 0.988 g/mL}
What is the molarity of CH3OHOH} in this solution?
#1 is answered above.
#2.
0.988 density = 0.988 g/mL.
So 1000 mL has a mass of 988 grams.
Only 6.4% of that is CH3OH so
988 x 0.0640 = 63 something is g CH3OH.
moles = 63/32 = about 2 M
You need to work through these estimates and use more accurate numbers.
To find the amount of grams needed for a sample known to be a certain percentage by mass, we can use the formula:
Mass of the sample = (Percentage/100) x Total mass
1.) For the first question, you need to find the grams of a sample known to be 99.81% AgNO3 by mass. Let's assume the total mass of the sample is "x" grams.
Mass of AgNO3 in the sample = (99.81/100) x x = 0.9981x
Now, since you are asked to prepare a solution of AgNO3 with a specific molarity, you will need to know the molar mass of AgNO3, which is approximately 169.87 g/mol.
To find the moles of AgNO3 required for the solution, we use the formula:
Moles = Molarity x Volume (in liters)
Given:
Molarity = 3.15x10^-2 M
Volume = 150.0 mL = 150.0/1000 = 0.150 L
Moles of AgNO3 = (3.15x10^-2) x (0.150) = 4.725x10^-3 mol
Now, using the formula:
Moles = Mass/Molar Mass
We can find the mass of AgNO3 required:
4.725x10^-3 mol = Mass/169.87 g/mol
Mass = 4.725x10^-3 x 169.87 = 0.801g
Therefore, you would need approximately 0.801 grams of AgNO3 for the solution.
2.) For the second question, you need to find the molarity of CH3OH in the solution, given that the solution is 6.40% methanol (CH3OH) by mass, and the density (d) is 0.988 g/mL.
First, we need to find the amount of methanol in the solution. Let's assume the total mass of the solution is 100 grams.
Mass of CH3OH in the solution = (6.40/100) x 100 = 6.40 grams
Now, to find the volume of the solution, we can use the formula:
Volume = Mass/Density
Given:
Mass = 100 grams
Density = 0.988 g/mL
Volume = 100/0.988 = 101.21 mL
Since the molarity is defined as moles of solute divided by the volume of the solution in liters, we need to convert the volume from mL to liters:
Volume = 101.21 mL = 101.21/1000 = 0.10121 L
Next, we need to know the molar mass of CH3OH, which is approximately 32.04 g/mol.
To find the moles of CH3OH in the solution, we use the formula:
Moles = Mass/Molar Mass
Moles of CH3OH = 6.40 g / 32.04 g/mol = 0.1999 mol
Finally, we can calculate the molarity by dividing the moles of CH3OH by the volume of the solution in liters:
Molarity = Moles/Volume
Molarity = 0.1999 mol / 0.10121 L = 1.972 M
Therefore, the molarity of CH3OH in the solution is approximately 1.972 M.