A beaker had 20 L of alcohol-glycerol mixture in the ratio 4 : 1 by volume. In the first round, 4 L of the mixture is removed and replaced with glycerol. In the second round, 5 L of the resultant solution is removed and replaced with glycerol. Finally, 10 L of the resultant mixture is removed and replaced with glycerol. What is the final quantity of glycerol in the mixture

• Here we are replacing the mixture with glycerol. So we have to take Alcohol concentrations for IC and FC.
  Initial concentration of alcohol is 4/5 = 80%
  Applying the formula for the first replacement:
  ⇒
  F
  C
  1
  =
  80
  %
  ×
  (
  1
  −
  4
  20
  )
  ⇒FC1=80%×(1−420)
  Here
  F
  C
  1
  FC1 is the concentration after first replacement.
  Second replacement:
  ⇒
  F
  C
  2
  =
  F
  C
  1
  ×
  (
  1
  −
  5
  20
  )
  ⇒FC2=FC1×(1−520)
  Third Replacement:
  ⇒
  F
  C
  3
  =
  F
  C
  2
  ×
  (
  1
  −
  10
  20
  )
  ⇒FC3=FC2×(1−1020)
  Now substituting the
  F
  C
  1
  FC1 and
  F
  C
  2
  FC2 in
  F
  C
  3
  FC3 we get
  ⇒
  F
  C
  3
  =
  80
  %
  ⇒FC3=80%
  ×
  (
  1
  −
  4
  20
  )
  ×(1−420)
  ×
  (
  1
  −
  5
  20
  )
  ×(1−520)
  ×
  (
  1
  −
  10
  20
  )
  ×(1−1020)
  
  ⇒
  F
  C
  3
  =
  80
  %
  ×
  4
  5
  ×
  3
  4
  ×
  1
  2
  ⇒FC3=80%×45×34×12
  
  ⇒
  F
  C
  3
  =
  24
  %
  ⇒FC3=24%
• 6 years agoHelpfull: Yes(0) No(1)
• primarily
  Alcohol & Gly= 16L & 4 L
  after 4L mixture replaced with glycrol:
  Alcohol= 16-4*4/5=64/5 L
  and Gly=20-64/5=36/5L
  and the new ratio=64/5:36/5=16:9
  
  Now after 5L mixture replaced with glycrol:
  Alcohol= 64/5-5*16/25=48/5 L
  and Gly=20-48/5=52/5 L
  and the new ratio=48:52=12:13
  
  after 10L mixture replaced with glycrol:
  
  Alcohol=48/5-10*12/25=24/5 L
  so Glycerol=20-24/5=76/5 =15.2 L
• 5 years agoHelpfull: Yes(0) No(0)