(1)
Given,
Mass concentration of A=2.59g/dm^-3
Mass of concentration B=mass/volume in
dm^3
=3.01/0.25 =12.04g/dm^-3
volume of base(VB)=25.0cm^3
volume of acid:(VA)=30.50cm^3
(1ai)
Molar mass of A(HCl)
=1+35.5=36.5g/mol^1 Concentration in mol/dm3=concentration in
g/dm3^1 /molar mass
=2.52/36.5
CA=0.069mol/dm^3
(1aii)
Concentration of B in g/dm^3 =mass in g/volume in dm^-3
=3.01/0.25
=12.04g/dm^3
(1aiii)
Mole of ratios of acid and base
nA=2, nB=1 Using; CAVA/CBVB=nA/nB
(0.069*30.50)/(CB*25.0)=2/1
CB:(0.069*30.50*1)/25.0*2
CB=2.1045/50
=0.042mol/dm^3
Concentration of B in mol/dm^3 is 0.042mol/dm^3
(1bi)
Molar mass of Na2Co3 . xH2O
= (23*2)+(12)+(16*3)+ x(1*2+16)
=46+12+48+18x
=106+18x 0.042=12.04/x
x=12.04/0.042=286.7gmol^-1
286.7=106+18x
18x=286.7 -106
18x=180.7
x=180.7/18 x=10
The number of molecules of water of crystallization present is 10.
- (2A)i) INFERENCE: Soluble salts
- (2A)ii) OBSERVATION: Whiteprecipitate
- (2A)iii) OBSERVATION: Insoluble in excess
- (2A)iv) OBSERVATION: insoluble excess
- INFERENCE: Na+ present
- (2b)i) INFERENCE: Zn^2+, Al^3+, Pb^2+ present
- (2b)ii) INFERENCE: Al^3+, Pb^2+ present
- (2B)iii) OBSERVATION: black
- precipitate is formed and precipitate dissolve
(3A)i) Measuring cylinder
(3A)ii) Kipp's Appararatus
(3A)iii) Separating funnel method
(3B) i. It turnsred litmus paper to blue
ii. colours remain unchanged
iii. It turns blue litmus paper red
(3C) i. AgNo3/NH3
ii. NaHCO3
iii. I2/NaoH
(3D) i. carbon(iv)oxide, carbon (ii)oxide
ii. Aluminium(III)oxide, Iron(III)oxide
MORE ANSWER LOADING.........
No comments:
Post a Comment