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
How many grams of potassium bromide are needed to prepare 300.0 mL of a solution that...
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