Question

1. Calulate the %(m/v) ad M of a solution made by dissolving .4g of K2SO4 in water and diluting to a volume of 50.00 ML.

2. Show (by use of eq. 5) how the first dilution of the 1 M solution of NaCl results in a .01 M Solution

3. How many mL of a stock solution of a 5.0% solution of NaNo3 are needed to make 1.5 L of a .25 % solution of NaNo3.

Please show work

Answer 1

1 ) We have ,

                Weight of K2SO4 ,Wt = 0.4 gr , V= 50ml or 0.05 L

     % (m /v ) = 0.4 / 0.05 * 100

                   = 800

M = Wt / Mwt * 1000/V(ml)

M = 0.4 / 174.2 * 1000 / 50

M = 0.046M

2 )    We have 1M Solution ; required one is 0.01M

     M1V1 = M2V2

    (1) V1 = (0.01) V2

        V2 / V1 = 1 / 0.01

        V2 / V1 = 100

To Dilute a solution from 1M to 0.01M , The solution is diluted to 100x it's original value .

3 ) We have ,

                  5.0% of stock solution of NaNO3 , Diluted to 0.25% of 1.5L

                  M1V1 = M2V2

                5 * V1   = 0.25 * 1.5

                    V1   = 0.25 * 1.5 / 5

                     V1   = 0.075 L OR 75 ml

75 ml of 5.0% stock solution is needed to make 1.5L of 0.25% solution of NaNO3