A 5 N solution is diluted 1:4. The resulting solution is diluted 4:15. What is the concentration in normality of the final solution?
5 * 1/4 * 4/15 = 5/15 = 1/3 N
What is the formula used here?
Well, let me calculate that for you.
So, we start with a 5 N solution and dilute it 1:4, meaning we mix 1 part of the 5 N solution with 4 parts of solvent. This would give us a new solution with a concentration of 1/5 of 5 N, which is 1 N.
Then, we further dilute this new solution 4:15, meaning we mix 4 parts of the 1 N solution with 15 parts of solvent.
Now, if you mix a clown with a lot of water, you get diluted clown juice. And if you mix that diluted clown juice with even more water, you get a lower concentration of clowniness.
So, in this case, the final solution would have a concentration of 4/15 of 1 N, which is approximately 0.267 N.
Hence, the concentration in normality of the final solution is about 0.267 N.
To find the concentration in normality of the final solution, we need to calculate the normality at each step of dilution and then multiply them together.
Step 1: Dilution 1:4
This means that for every 1 part of the 5 N solution, we are adding 4 parts of the solvent (usually water).
To calculate the normality after the first dilution, we can use the formula:
Normality1 = Normality0 * (Volume0 / Volume1)
Normality0 = 5 N (initial normality)
Volume0 = 1 part of the 5 N solution
Volume1 = 1 part of the 5 N solution + 4 parts of the solvent
Plugging in the values:
Normality1 = 5 N * (1 / (1 + 4))
Simplifying:
Normality1 = 5 N * (1 / 5)
Normality1 = 1 N
Step 2: Dilution 4:15
This means that for every 4 parts of the first diluted solution, we are adding 15 parts of the solvent (usually water).
To calculate the normality after the second dilution, we use the same formula as above:
Normality2 = Normality1 * (Volume1 / Volume2)
Normality1 = 1 N (normality after the first dilution)
Volume1 = 4 parts of the first diluted solution
Volume2 = 4 parts of the first diluted solution + 15 parts of the solvent
Plugging in the values:
Normality2 = 1 N * (4 / (4 + 15))
Simplifying:
Normality2 = 1 N * (4 / 19)
Therefore, the concentration in normality of the final solution is approximately 0.211 N.
To find the concentration in normality (N) of the final solution, we need to understand the concept of normality and how it is affected by dilution.
Normality (N) is a unit of concentration used in chemistry that measures the equivalent concentration of a solute in a solution. It is defined as the number of equivalent weights of solute per liter of solution. The formula to calculate normality is:
Normality (N) = (number of equivalents of solute) / (volume of solution in liters)
In this problem, we have a 5 N solution that is diluted 1:4. This means that the solution is being mixed with 4 parts of a solvent for every 1 part of the original solution.
To dilute a solution, we can use the following equation:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Let's calculate the concentration (C2) and volume (V2) after the first dilution:
C1 = 5 N
V1 = ?
C2 = ?
V2 = 4 parts (since the solution is diluted 1:4)
Using the dilution equation, we can rearrange it to solve for C2:
C2 = (C1 * V1) / V2
Since we don't know the initial volume (V1), we'll use a variable to represent it. Let's use "x".
C2 = (5 N * x) / 4
Now, we will further dilute the resulting solution with a ratio of 4:15. This means that it is being mixed with 15 parts of a solvent for every 4 parts of the previous solution.
Let's calculate the final concentration (C_final) and final volume (V_final):
C2 = ?
V2 = ?
C_final = ?
V_final = 15 parts
Using the dilution equation again:
C2 * V2 = C_final * V_final
Substituting the values we have:
((5 N * x) / 4) * V2 = C_final * (15 parts)
Simplifying the equation, we have:
C_final = ((5 N * x) / 4) * (4 / 15)
C_final = (5 N * x) / 15
So, the concentration in normality of the final solution is (5 N * x) / 15, where "x" represents the initial volume. Since the information about the initial volume is not provided, we cannot determine the exact concentration in normality of the final solution without this information.