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?
M = moles/L
You know M and L, calculate moles.
Then moles = g/molar mass. You know moles and molar mass, calculate g AgNO3 (theoretical AgNO3) but since it isn't 100%, you must correct for that.
grams to weigh *0.9981 = theoretical AgNO3. Solve for grams to weigh.
thank you
To answer the first question, you need to determine the mass of the AgNO3 sample that would contain 99.81% AgNO3 by mass.
1. Start by assuming that the mass of the sample is 100 grams (a convenient assumption).
2. Since the sample is known to be 99.81% AgNO3 by mass, you can calculate the mass of AgNO3 in the sample by multiplying the total mass of the sample by the mass percentage of AgNO3:
Mass of AgNO3 = (99.81 / 100) * 100 grams
3. Now that you know the mass of AgNO3, you can convert it to grams.
For the second question, you are given the mass percentage of methanol (CH3OH) in the solution and its density. You need to determine the molarity of CH3OH in the solution.
1. Start by assuming that you have 100 grams of the solution (a convenient assumption).
2. Calculate the mass of CH3OH in the solution by multiplying the total mass of the solution by the mass percentage of CH3OH:
Mass of CH3OH = (6.40 / 100) * 100 grams
3. Now that you know the mass of CH3OH, you can convert it to moles using the molar mass of CH3OH.
To calculate the molarity of CH3OH in the solution:
1. Convert the mass of CH3OH to moles using the molar mass of CH3OH.
2. Calculate the volume of the solution (in liters) using its density.
3. Divide the moles of CH3OH by the volume of the solution to get the molarity.