A 45.8mL sample of 5.8M KNO3 solution is diluted to 1.00L. What volume of the diluted solution contains 15.0 g of KNO3.
Chemically
Well, when it comes to diluting solutions, it's like adding a little water to that super strong punch at a party. So, let's do a little math circus performance here!
First, we need to find out how much KNO3 is in that 45.8 mL of 5.8M solution. We can use the equation:
moles = Molarity × volume
Since we know the molarity (5.8M) and the volume (45.8 mL), we can plug those numbers in to calculate the number of moles.
moles = 5.8M × 0.0458L (since we converted mL to L)
Solving that equation will give us the number of moles of KNO3 in the original solution.
Now, we need to figure out how many moles of KNO3 we want in the diluted solution. To do that, we need to use the equation:
moles = mass / molar mass
We know the mass of KNO3 is 15.0g, and the molar mass of KNO3 is 101.11 g/mol. Plugging in these numbers will give us the number of moles of KNO3 we want.
Finally, we can find the volume of the diluted solution that contains 15.0g of KNO3. We'll use the equation:
volume = moles / Molarity
Plugging in the number of moles we want and the molarity of the diluted solution (which is 5.8M since it remains the same) will give us the volume in liters.
So, if you give me a moment, I'll calculate that for you. *starts juggling numbers*
*calculates furiously*
Okay, I've got it! The volume of the diluted solution that contains 15.0g of KNO3 is approximately 0.045 L or 45 mL. So, you'll need around 45 mL of this diluted solution to get your desired amount of KNO3.
And remember, my calculations are like a comedy routine - sometimes they're accurate, and sometimes they're just for laughs!
To find the volume of the diluted solution that contains 15.0 g of KNO3, we first need to determine the concentration of the diluted solution after dilution.
Given:
Initial volume (V1) = 45.8 mL
Initial concentration (C1) = 5.8 M
Final volume (V2) = 1.00 L
Final concentration (C2) = ?
To find the final concentration (C2), we can use the formula:
C1V1 = C2V2
Substituting the given values, we have:
5.8 M × 45.8 mL = C2 × 1000 mL (since 1 L = 1000 mL)
Simplifying the equation:
C2 = (5.8 M × 45.8 mL) / 1000 mL
C2 = (5.8 × 45.8) / 1000
C2 = 0.26564 M
Now, we can use the final concentration (C2) to calculate the volume of the diluted solution containing 15.0 g of KNO3.
Using the formula for concentration (C):
C = n / V
Where:
C is the concentration,
n is the amount of solute (in moles),
V is the volume of the solution (in liters).
Since we want to find the volume (V), we can rearrange the formula:
V = n / C
We need to determine the number of moles (n) of KNO3 in 15.0 g. To do this, we can use the molar mass of KNO3, which is the sum of the atomic masses of each element in the compound.
The molar mass of KNO3 is:
Mass of K = 39.1 g/mol
Mass of N = 14.0 g/mol
Mass of O (3) = 16.0 g/mol
Total molar mass of KNO3 = (39.1 g/mol) + (14.0 g/mol) + (16.0 g/mol × 3) = 101.1 g/mol
Next, we can calculate the number of moles (n) of KNO3:
n = mass / molar mass
n = 15.0 g / 101.1 g/mol
n = 0.148 mol
Finally, we can find the volume (V) of the diluted solution that contains 15.0 g of KNO3:
V = n / C2
V = 0.148 mol / 0.26564 M
V ≈ 0.557 L
Therefore, approximately 0.557 liters (or 557 mL) of the diluted solution will contain 15.0 g of KNO3.
To find the volume of the diluted solution that contains 15.0 g of KNO3, you need to first calculate the concentration of the diluted solution.
To dilute the 45.8 mL of 5.8M KNO3 solution to 1.00L, you need to add water. The amount of water added is given by the difference between the final volume and the initial volume:
Volume of water added = Final volume - Initial volume
The final volume is 1.00L, and the initial volume is 45.8 mL. We need to convert the initial volume to liters before performing the calculation:
Initial volume = 45.8 mL = 45.8 mL * (1 L / 1000 mL) = 0.0458 L
Volume of water added = 1.00 L - 0.0458 L = 0.9542 L
To calculate the concentration of the diluted solution, we need to consider that the moles of KNO3 remain constant in the dilution process. The moles of KNO3 in the initial solution can be calculated using the formula:
moles = volume (L) * concentration (M)
In the initial solution, the moles of KNO3 are given by:
moles of KNO3 = 0.0458 L * 5.8 M = 0.26564 mol
Since moles of KNO3 are constant, the same amount is present in the final diluted solution, which means:
moles of KNO3 in the diluted solution = 0.26564 mol
To find the concentration of the diluted solution, we rearrange the formula:
concentration (M) = moles / volume (L)
The volume of the diluted solution that contains 15.0 g of KNO3 can be calculated by rearranging the formula again:
volume (L) = moles / concentration (M)
Substituting the given values:
volume (L) = 15.0 g / (0.26564 mol / 1.00 L) = 56.424 L
Therefore, the volume of the diluted solution that contains 15.0 g of KNO3 is 56.424 L.
What's the M of the diluted solution? That will be 5.8M x (45.8/1000) = about 0.26M
How many mols in 15.0 g KNO3? That is 15.0/molar mass KNO3 = about 0.15 moles.
So what volume of the diluted solution do we need to make 0.15 moles.
M = moles/L
0.26 M = 0.15 moles/L
L = about 600 mL.
Check my work. It's late and past time for my bed.