A sample of 0.8103 g of KMnO 4 was dissolved in water and made up to the volume
ID: 540061 • Letter: A
Question
A sample of 0.8103 g of KMnO4 was dissolved in water and made up to the volume in a 500.0mL volumetric flask. A 2.000mL sample of this solution was transferred to a 1000mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0mL flask and diluted to the mark with water.
(a) Calculate the concentration (in molarity) of the final solution. Enter your answer in scientific notation.
(b) Calculate the mass of KMnO4, needed to directly prepare the final solution. Enter your answer in scientific notation.
Explanation / Answer
m = 0.8103 g of KMNO4
V = 500 mL
C1 = m/V= 0.8103 /500 = 0.0016206 g/mL
then...
we take V = 2 mL, C = 0.0016206 g/mL
dilution up to 1000 mL
Cnew = v1/v2*C = 2/1000*0.0016206
Cnew = 0.0000032412 g/mL
then
v = 10 mL are diluted to 250 mL
Cnew = vold/vnew * C = 10/250 * 0.0000032412 =1.29648*10^-7 g/mL
b)
mass of KMnO4 required for final dilution
V = 250 mL
mass = C*V = 250*(1.29648*10^-7) = 0.000032412 g of KMnO4 required