Calculate the normality of a 200 ml of H2SO4 solution which when treated with excess of BaCl2 solution precipitates 2.33 g of BaSO4.
Answers
Answer:
ormality of any acid or base will be equal to gram-equivalent desolve in given volume
N = gram-equivalent/ volume(lt.)
When two or more compounds or elements react together their
gram-equivalent will be equal in simple way, 10 gm-eq of an acid will react completely with 10 gm-eq of base,
In given problem,
gm-eq. of H2SO4 = N*V(lt.)
and N = n*M where, M = molarity and n = basicity or acidity of an acid or base.
gm-eq of H2SO4 = 100*2*0.2*0.001
=0.04gm-eq
gm-eq of NaOH = 100*0.2*1*0.001
=0.02
remaining gm-eq of H2SO4 = 0.04–0.02
0.02gm-eq
Now,
gm-eq. = N*V = 0.02
Total volume =200ml = 0.2 lt.
N*0.2 = 0.02 = 0.1N
Answer:
From following relation we can say that
137 g Ba. > 208g BaCl2
X. g Ba > 2.08 g Bacl2
This show that 208g Bacl2 containing 137g Ba so that 2.08 g BaCl2 containing Xg Ba
Hence X ×208 = 137×2.08
X. = 137×2.08 ÷ 208 g Ba
X= 1.37 g present in given 2.08. g BaCl2 solution
Also 137g Ba = 233 g. BaSO4
1.3 7. g Ba. = X g BaSO4
137 g Ba correspond with 233g BaSO4 .1.37 g Ba. Corr with Xg BasO4
X × 137 = 233×1.3 7
X = [233×1.37]÷137 g BaSO4
X= 2.33 g of. BaSO4
Hence 2.33 g of BaSO4 will be forme during performing this experiment.