Posted by Merlina_Lyn on Wednesday, January 30, 2013 at 10:18am.
I would help, but I am not all that positive about the information provided. This is what I believe you are saying: You want to make 200mL of a 0.4M solution using 25% ammonia by m/v. However, I do not know if the value for M2 and V1 are values that you calculated or are values that are provided for you as the answer, but if they are the answers that you calculated, I do not believe they are correct.
25% of ammonia by m/v= 25g/mL
25g/17.03 g of NH3 *mole-1=1.47 moles of NH3
Therefore, the 25% of ammonia by m/v =1.47 moles/mL
In order to use the above equation, which is correct, I need to make sure that my units are the same. It is easier to convert mL to L, so I will do that for the initial solution and for the volume of 0.4M solution needed.
1.47moles/mL*(1mL/10^-3 L) =1470 moles/L=1470M
200mL*(10^3 L/1mL)=0.200L
I know that the molarity that I want is 0.4M, and I know the volume needed, is 0.200L. Since I already have converted my 25% solution to molarity, all I need to do is solve for the volume of the 25% solution that I need to dilute to make a 0.4M solution using the equation that you provided.
M1=1470M
V1= ?
M2=0.4M
V2=0.200L
M1V1=M2V2
(1470M)V1=(0.4M)(0.200L)
Solving for V1,
V1=[(0.4M)(0.200L)]/(1470M)=5.45 x 10^-5 L
5.45 x 10^-5 L*(10^3 mL/1L)= 5.45 x 10^-2 L
I hope this helps.
Last line should say 5.45 x 10^-2 mL