You dissolve 6.55 g of potassium nitrate in enough water to make a 250. ML solution.
1.What is the molarity of this solution?answer:0.0262
2.If you took out 15.0 mL of this solution, how many moles of potassium nitrate would you have?answer:0.393
3. What is the percent (m/v) of this solution?answer:0.0262
4.If the resulting solution has a density of 1.45 g/mL, what is the percent (m/m) of the solution?answer:0.0181
5.How does the freezing point of this solution compare with that of pure water?answer:should be lower
6.How does the boiling point compare?answer:it should be higher than that for pure water
7.If you added 1.25 L of additional water to this solution, what would the molarity of this new solution be?answer:.0432
4,5,6 look ok.
1,2,and 3 I can't confirm.
Show your work on those.
I didn't check 7.
This is what I did..please check it
1.P1V1=P2V2
P=6.55/250=.0262
2.V=P*V1/V2
P=6.55*15.0/250=.393
3.P=PV2V
PP=6.55/250=.0262
1, 2, and 3 are NOT gas law problems. They are concentration problems.
1.
mols = grams/molar mass
Then M = mols/L soln
2.
mols = M x L = ?
3. You have 6.55 g in 250 mL. Convert this to g/100 for % m/v.
DID I DO IT RIGHT NOW
1.M=6.55*1mol/101gn=.065
.065/250=.260
2.m=6.55g*15.0=.89 so this mean I need to use the full 250ml
3.6.55/250*100=2.62
correctly for #2 is 98.25
1. My calculator reads 0.25941 which I would round to 0.259 and not 0.260. The reason you obtained 0.260 is because you rounded incorrectly for the first part of the question; i.e., 6.55/101 = 0.06485. If you round that off too much that makes all of the remaining answers that follow that wrong. This is what you should do.
6.55/101 = some number BUT LEAVE THAT IN THE CALCULATOR. I don't even read the number and that saves some time. Then divide by 0.250L. That way you don't make rounding errors. ;-).
2.
The question is to find the number of mols in a 15.0 mL portion.
mols = M x L = 0.0259M x 0.015L = ?
3. 2.62 is correct but some profs get picky and will count off if you don't write the unit, also. I would write 2.62% m/v
7. Now that you have #1, #7 follows:
0.0259M x [250 mL/(250 mL + 1250 mL)] = ? or approximately 0.004 M
To calculate the molarity of a solution, you need to know the number of moles of solute and the volume of the solution in liters. Here's how you can calculate the molarity for the given solution:
1. We are given that 6.55 g of potassium nitrate is dissolved in enough water to make a 250 mL solution. First, convert the volume from milliliters to liters by dividing it by 1000: 250 mL ÷ 1000 = 0.25 L.
2. Next, calculate the number of moles of potassium nitrate using its molar mass. The molar mass of potassium nitrate (KNO3) is approximately 101 g/mol. To find the number of moles, divide the mass of potassium nitrate (6.55 g) by its molar mass: 6.55 g ÷ 101 g/mol = 0.0649 mol.
3. Finally, calculate the molarity by dividing the number of moles by the volume in liters: molarity = 0.0649 mol ÷ 0.25 L = 0.2596 M or approximately 0.0262 M (rounded to four decimal places).
So, the molarity of this solution is approximately 0.0262 M.
To answer the second question, if you were to take out 15.0 mL of this solution, you need to calculate the number of moles of potassium nitrate present in that volume:
1. First, convert the volume from milliliters to liters by dividing it by 1000: 15.0 mL ÷ 1000 = 0.015 L.
2. Use the molarity calculated in the previous question (0.0262 M) to calculate the number of moles of potassium nitrate in the given volume. Multiply the molarity by the volume: 0.0262 M × 0.015 L = 0.000393 moles or 0.393 moles.
So, if you took out 15.0 mL of the solution, you would have approximately 0.393 moles of potassium nitrate.
To calculate the percent (m/v) of the solution, you need to divide the mass of the solute (potassium nitrate) by the volume of the solution and then multiply by 100:
1. The mass of the solute is given as 6.55 g.
2. The volume of the solution is given as 250 mL. However, it is good practice to convert this volume to liters to ensure consistency. Convert the volume from milliliters to liters by dividing it by 1000: 250 mL ÷ 1000 = 0.25 L.
3. Calculate the percent (m/v) by dividing the mass of the solute by the volume of the solution and multiplying by 100: percent (m/v) = (6.55 g ÷ 0.25 L) × 100 = 26.2%.
So, the percent (m/v) of this solution is 26.2%.
To calculate the percent (m/m) of the solution, you need to divide the mass of the solute (potassium nitrate) by the mass of the solution and then multiply by 100:
1. The mass of the solute is given as 6.55 g.
2. The total mass of the solution can be calculated by multiplying the volume of the solution (250 mL) by its density (1.45 g/mL): 250 mL × 1.45 g/mL = 362.5 g.
3. Calculate the percent (m/m) by dividing the mass of the solute by the mass of the solution and multiplying by 100: percent (m/m) = (6.55 g ÷ 362.5 g) × 100 = 1.81%.
So, the percent (m/m) of this solution is 1.81%.
The freezing point of the solution should be lower than that of pure water. This is because the presence of solute particles (potassium nitrate) in the solution lowers the freezing point of the solvent (water). This phenomenon is known as freezing point depression.
The boiling point of the solution should be higher than that of pure water. This is due to the same reason as above; the presence of solute particles in the solution raises the boiling point of the solvent.
To calculate the molarity of the new solution when 1.25 L of additional water is added, you need to consider that the amount of solute (potassium nitrate) remains the same:
1. The volume of the new solution would be the sum of the initial solution volume (0.25 L) and the volume of additional water added (1.25 L): 0.25 L + 1.25 L = 1.5 L.
2. The number of moles of potassium nitrate present in the new solution will remain the same as before (0.0649 mol).
3. Calculate the new molarity by dividing the number of moles by the new volume: molarity = 0.0649 mol ÷ 1.5 L = 0.0864 M or approximately 0.0432 M (rounded to four decimal places).
So, the molarity of the new solution would be approximately 0.0432 M.