What weight of hydrated oxalic acid should be added for complete neutralisation of 100 mL of 0.2 N -NaOH solution?
0.45 gm
0.90 gm
1.08 gm
1.26 gm
D.
1.26 gm
For complete neutralisation of 100 mL of 0.2 N NaOH solution, 100 mL of 0.2 N hydrated oxalic acid solution is required.
This is due to the fact that at neutral point gram equivalent of an acid is equal to gram equivalent of base.
Normality =
where, a = Weight of solute
E = equivalent weight of solute
b = Volume of solution
0.2 =
(Equivalent weight of hydrated oxalic acid is 63)
a = = 1.26 gm
Hence, weight of hydrated oxalic acid is 1.26 gm.
The number of moles of a solute in its solution is 20 and total number of moles are 80. The mole fraction of solute is
2.5
0.25
1
0.75
The relation between Kp and Kc is correctly shown as
Kc = Kp (RT)
Kp = Kc (RT)
Kp = Kc (RT)
Kc = Kp (RT)
In the formation of SO3 according to following equilibrium 2SO2 + O2 2SO3 + heat, which conditions favour the formation of SO3?
Low pressure
Low concentration of oxygen
Low temperature and use of catalyst
High temperature without using catalyst
Which of the following mixtures forms an acid buffer?
NaOH + HCl
CH3COOH + CH3COONa
NH4OH + NH4Cl
H2CO3 + (NH4)2CO3
When NH4Cl is added to NH4OH solution, the dissociation of ammonium hydroxide is reduced. It is due to
common ion effect
hydrolysis
oxidation
reduction
pH of a solution is 4. The hydroxide ion concentration of the solution would be
10-4
10-10
10-2
10-12
Heat of neutralisation of NH4OH and HCl is
equal to 13.7 kcal
more than 13.7 kcal
less than 13.7 kcal
more than one is correct