The solution with highest normality is
O 4.9%
H,SO
08g NaOH in 2lit
43 milli equivalents in 100ml
| 1M H P0 ,
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No of mili Equivalents = Molarity × n-factor × volume(mili liters)
Or No of mili equivalents is also = normality × volume ( mili liters)
No of equivalents of Base added = 0.5 × 100 = 50
No of equivalents of acid = 3 × 10 + 1 × 20 = 50 ( since two acids are added so we have to calculate total equivalents of both acids )
Since milliequivalents of acid and base are equal hence the solution is neutral.
Option C is correct.
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