For preparation of 4N solution of sulphuric acid
Dissolve 196 gm H2SO4 in 1000 ml DM water.
Calculation
N E V
Weight = ------------
1000
Where
N= Normality
E= Equivalent weight
V= Volume
normality of A = #gram equivalent weights A divided by #litres of solution
thus: measure a gram equivalent of H2SO4 (49g) and add to 1000ml volumetric flask that has 500ml deionized distilled H2O. stopper flask, mix by swirling, open flask and fill to 1000ml mark, thoroughly mix
OR
1N H2SO4 = 0.5M H2SO4
volumetrically measure 500ml 1M H2SO4
volumetrically transfer to 1000ml volumetric flask (using DDH2O)
fill to 1000ml mark with DDH2O
mix
OR
1) Measure 28ml Sulphuric Acid (95% reagent grade)
2) Transfer all into voulmetric flask (contain 500ml Carbon dioxide free distilled water)
3) Mix by swirling
4) Fill up to 1000ml mark with CO2 free distilled water.
Sulphuric acid Equivalent weight = 49
Normality you can decided i.e. you prepare 0.1N
water 1 lit of water
First = Eq.wt x 0.1 N x 1 lit of water
= 49 x 0.1 x 1= 4.9 g
Sulphuric acid you only measure only ml
so you put = gram / (Specific Gravity x Percentage Concentration)
= 49 / (1.83x97)
= 49/177.51
= 2.76 ml of Sulphuric acid you take one litre of water
0.01 N sulphuric acid means 0.01 equivalents dissolved in 1 litre of water.
Nomality = Molarity x n-factor
=> molarity= normality /n-factor
molarity = 0.01/2=0.005 M
[ n-factor of sulphuric acid is 2]
Thus, a 0.01N solution of sulphuric acid contains 0.005 moles of sulphuric acid in 1 litre.
0.005 moles=0.005 x 98 = 0.49 grams.
=> Dissolve 0.49 grams of sulphuric acid in 1 litre water to make a 0.01N solution
The normality of sulphuric acid in the bottle is 36N
But we dont know how much ml we have to take from the bottle.
to make clear go through the calculation....
preparing 0.01 N in 1000ml
X * 36N =0.01 *1000ml
x=0.27 ml (but it is very difficult to measure,,,so you first prapare 10N of sulphuric acid
X * 36N = 10N *100ML
X=27.7 ml of sulphuric acid and add 73.3 ml of water to makeup for 100ml(10N SULPHURIC ACID)
FROM the 10N
X* 10N = 0.01N *1000
X=1ml of sulphuric acid mixed in 999ml of water
Add one mole of sulphiric acid to one liter of water.
The concentrated available has a normality of 36. so when you need to prepare 0.01 N Sulfuric acid calculate accordingly.
20 ml 2 normal dilute sulphuric acid means?
7 N h2s04
0.49
how do prepare 0.1 N Oxalic acid
sulphuric acid
Yes
Yes
Benzoic acid and n-octyl alcohol in presence of sulphuric acid on heating form octyl benzoate and water.
yes
how do prepare 0.1 N Oxalic acid
sulphuric acid
Yes
Yes
when we add water n sulphuric acid then dilute sulphuric acid is formed. But we have to mix concentrated acid to water not water to acid otherwise the container in which u r mixing may explode because this reaction is highly exothermic.
Yes
Benzoic acid and n-octyl alcohol in presence of sulphuric acid on heating form octyl benzoate and water.
dilute with water
7n h2so4
Different types of urine preservatives are boric acid, tartaric acid, thymol, clorhexidine/ n-propyl gallate and sulphuric acid.
Concentration of a solution is calcuated by dividing the number of moles by the volume. C = n/v.