1. What molarity is needed so that 5mL diluted to 475 mL will yield 0.1M?
For this one I used M1V1=M2V2 and got 9.5M just checking to see if I did that right.
2. How much 2.5 HNO3 can be made from 7.0mL of a 16.0M HNO3 soltuion?
I tried using M1V1=M2V2 for this one too but wound up needing 44.8mL. I can't think of another way to try to solve the problem.
1. the solution is diluted by 5/475 dilution factor i.e. 1/95x
so you can also use this factor to find the initial molarity (M1) by dividing the final molarity (0.1M) with 1/95..i.e. 0.1x95 = 9.5M.
2. c1v1=c2v2 and we are looking for v2 so;
v2=(c1v1)/c2 = (7x16)/2.5 = 44.8mL
i got the same answer with you..if you are in doubt, then you can calculate the dilution factor and use it to determine the final concentration;
so, initial volume is 7mL and final is 44.8mL; dilution factor = 7/44.8 = 0.15625x
multiply this with the initial concentration and see what happens...
if it gives 2.5M then 44.8mL is correct...
hope that helps..
Oh I think I was just looking at the question the wrong way. I was wondering how I could get 44.8mL from 7.0mL forgetting that the 2.5M solution would need to be diluted. Thanks!
1. To calculate the molarity needed to achieve a desired concentration after dilution, you correctly used the equation M1V1 = M2V2. Let's go through the calculations to confirm if you arrived at the correct answer.
Given:
V1 (initial volume) = 5 mL
M2 (desired molarity after dilution) = 0.1 M
V2 (final volume) = 475 mL
Substituting the values into the equation, we have:
M1 (initial molarity) * V1 = M2 * V2
M1 * 5 mL = 0.1 M * 475 mL
Next, solve for M1:
M1 = (0.1 M * 475 mL) / 5 mL
M1 = 4.75 M
So, the correct molarity needed to achieve a 0.1 M concentration after dilution is approximately 4.75 M. Therefore, your calculation of 9.5 M appears to be incorrect.
2. To determine how much 2.5 M HNO3 can be made from a 16.0 M HNO3 solution, you correctly used the equation M1V1 = M2V2. Let's walk through the calculations to verify your answer.
Given:
V1 (initial volume) = 7.0 mL
M1 (initial molarity) = 16.0 M
M2 (desired molarity) = 2.5 M
Substituting the values into the equation, we have:
M1 * V1 = M2 * V2
16.0 M * 7.0 mL = 2.5 M * V2
Next, solve for V2:
V2 = (16.0 M * 7.0 mL) / 2.5 M
V2 ≈ 44.8 mL
So, you arrived at the correct answer. Approximately 44.8 mL of a 16.0 M HNO3 solution is needed to produce 2.5 M HNO3. Well done!