Question

How many grams of potassium bromide are needed to prepare 300.0 mL of a solution that is 2.0%w/v? Do NOT include any units, just answer with the NUMBER of grams, for example: 4.2 or 13 or 220. Type your response

Answer 1

1) To prepare 300 mL of 2.0 % w/v KBr

    2.0 % w / v means that 100 mL of solution has 2 g of KBr.

Therefore 300 mL of solution would have 300 x 2 / 100 = 6.0 g KBr.

Thus the grams of KBr needed to prepare 300 mL of 2.0% w/v is = 6.0

2)

      20 mL of 4% w/v solution has [20 x 4 / 100] g of solute.

                                                   = 0.8 g of solute.

When solvent is added and the new volume = 100 mL

mass of solute - 0.8 g

Volume of solution = 100 mL

Therefore w/v = 0.8/100

             % w/v = [0.8 / 100] x 100 = 0.8 %

[Alternatively use V1C1 = V2C2,

V1 = 20 mL C1= 4 % w/v

V2 = 100 mL C2 = ?

C2 = V1C1 / V2 = 20 x 4 / 100 = 0.8 % w/v ]

Thus the new concentration is = 0.8

3)

V1M1 = V2M2

V1 = 25 mL M1 = 3.0 M

V2 = 100 mL M2 = ?

M2 = V1M1 / V2

      = 25 x 3.0 / 100

      = 0.75 M

Thus the concentration of new solution is = 0.75