MATHEMATICS ASSIGNMENT
QUESTION
Thirty liters of a 40% acid solution is obtained by mixing a 25% acid solution with a 50% acid solution. How much of each solution XX required XX obtain XXX XXXXX mixture?
XXXXXX
Given
Volume XX XXXXX XXXX XXXXXXXX = X = XX X
Concentration of XXXXX acid solution = X = 40% X/X
Concentration XX XXXXX XXXXXXXX = XX = 25% X/X
XXXXXXXXXXXXX XX XXXXXX solution = XX = 50% V/X
XX find
Volume of first solution XXXXXXXX = V1 =?
XXXXXX of XXXXXX XXXXXXXX required = V2 =?
XXXXXXXX
XX know that,
X = XX + V2
30 L = V1 + V2
V1 = XX – XX
XX XXXX first XXXX XXX XXXXXX XX XXXXXX in final solution,
Concentration (in XXXXXXXXXX) = (Amount XX XXXXXX / XXXXXX XX solution) × 100
C = (N / V) × XXX
XX% = (N / 30) × XXX
0.X = N / 30
N = 30 × 0.4
N = XX X
Now,
XXXXXX of solute in XXXXX XXXXXXXX = Amount of XXXXXX in V1 XX XXXXX XXXXXXXX + Amount of solute in V2 XX XXXXXX solution
N = N1 + XX
XX X = [(C1 / XXX) × XX] + [(XX / XXX) × XX]
12 = [(XX / XXX) × XX] + [(50 / 100) × 100]
12 = (0.25 × XX) + (0.5 × V2)
XXXXXXX the XXXXX XX XX,
XX = [0.25 × (30 – XX)] + (0.5 × XX)
12 = 7.5 – X.25V2 + 0.5V2
12 – X.X = 0.XXX – 0.XXXX
X.5 = 0.XXXX
V2 = 4.5 / X.25
XX = XX X
XX XXXX XX,
V1 = 30 – XX
V1 = 30 – XX
XX = 12 X
Result
XX = 12 L
V2 = XX X
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