1
Question
A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is
A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is
Open in App
Solution
Number of equivalents of the acid = Number of equivalents of the base
Equivalents of NaOH solution = volume × molarity
= 0.02×0.1=0.002
Oxalic acid dihydrate eq in 20 mL = 0.002 eq
Oxalic acid dihydrate eq in 100 mL = 0.01 eq
Oxalic acid dihydrate in the original solution = 0.01×10=0.1 eq
Mass of the acid = equivalents × equivalent mass
= 0.1×63=6.3 g
Number of equivalents of the acid = Number of equivalents of the base
Equivalents of NaOH solution = volume × molarity
= 0.02×0.1=0.002
Oxalic acid dihydrate eq in 20 mL = 0.002 eq
Oxalic acid dihydrate eq in 100 mL = 0.01 eq
Oxalic acid dihydrate in the original solution = 0.01×10=0.1 eq
Mass of the acid = equivalents × equivalent mass
= 0.1×63=6.3 g
Suggest Corrections
0
Similar questions
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
