Asked by Anonymous
22 ml nitric acid neutralises 20 ml , 0.5 normal sodium hydroxide sollution. calculate weight of nitric acid in 500 ml?
Answers
Answered by
Graham
For both nitric acid and sodium hydroxide, the acid hydrogen equivalence factor is 1. That is: 1N=1M.
Given:
22mL HNO3 neutralises 20mL, 0.5N NaOH, via the stoichiometry:
HNO3 + NaOH = NaNO3 + H2O
Molar Mass of HNO3 is 63.01296±0.00004 g/mol.
Let: c be the concentration of HNO3
Find: m, the mass of HNO3
.: (0.022L)c = (0.020L)(0.5M)
m = (0.500L)(63.01296g/mol)(0.020L)(0.5mol/L)/(0.022L)
m = 14.3211(27...) g
m ≈ 14 g
Given:
22mL HNO3 neutralises 20mL, 0.5N NaOH, via the stoichiometry:
HNO3 + NaOH = NaNO3 + H2O
Molar Mass of HNO3 is 63.01296±0.00004 g/mol.
Let: c be the concentration of HNO3
Find: m, the mass of HNO3
.: (0.022L)c = (0.020L)(0.5M)
m = (0.500L)(63.01296g/mol)(0.020L)(0.5mol/L)/(0.022L)
m = 14.3211(27...) g
m ≈ 14 g
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