calculate the 0.4M molarity of ammonia
From data given, how to prepare 200ml of 0.4M ammonia solution 25%. From previous study, i found that the calculation is M1V1=M2V2
From formula, Initial concentration (M1)is 910000mg/L, vOlume of beaker (V2)=200ml, but the required concentration (M2)? from previous study state M2=6815.2 mg/L <--------how to get this?? and V1 will be 1.498ml..do help me please
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
To calculate the required concentration (M2) for the 0.4M ammonia solution, you can use the formula M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the required concentration, and V2 is the final volume.
In this case, you have an initial concentration of 910,000 mg/L (0.4M) and a final volume of 200 ml. You need to calculate the required concentration (M2) to prepare a 0.4M ammonia solution.
First, convert the final volume into liters: V2 = 200 ml = 0.2 L.
Now, substitute the values into the formula: (M1)(V1) = (M2)(V2).
(0.4M)(V1) = (M2)(0.2 L)
Rearrange the formula to solve for M2: M2 = (0.4M)(V1) / (0.2 L).
We still need to know the value of V1, which is the initial volume. According to your calculations, V1 is 1.498 ml.
Now, substitute the value of V1 into the formula: M2 = (0.4M)(1.498 ml) / (0.2 L).
To convert milliliters to liters: 1 ml = 0.001 L. Therefore, V1 can be expressed as 1.498 ml * 0.001 L/ml = 0.001498 L.
Substitute the value of V1 into the formula: M2 = (0.4M)(0.001498 L) / (0.2 L).
Now, calculate M2: M2 = 0.0029968 M = 2996.8 mg/L.
So, the required concentration (M2) for preparing a 0.4M ammonia solution with a final volume of 200 ml is approximately 2996.8 mg/L.
Please note that this calculation assumes that the initial concentration given is correct and that all necessary units have been converted accurately. Always double-check the accuracy of your measurements and calculations before proceeding with the preparation of solutions.