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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 help me please

  • Chemistry -

    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.

    V1= ?


    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.

  • Chemistry -

    Last line should say 5.45 x 10^-2 mL

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