Sun 01 Apr 2018
PLS NOTE THAT YOU ARE REQUIRED TO USE UR SCHOOL'S TITRE VALUE IN NUMBER 1.
OUR SCHOOL GOT 15.30CM3.
REPLACE WITH UR SCHOOLS OWN AND RECALCULATE.
Volume of pipette used=25.0cm^3
Final reading: 16.00|15.40|17.20|15.30
Initial reading: 0.00|0.00 |2.00 |0.00
Vol of acid used: 16.00|15.40|15.20|15.30
Average volume of A used = (VA1 + VA2 + VA3) cm³
VA = (15.40 + 15.20 + 15.30)/3cm³
VA = 45.90/3 = 15.30cm³.
B contains 4.80g/250cm³
Hence x8 = 1000cm³
X = 4.80 × 1000/250 = 4.80 × 4 = 19.2gdm³
Hence molar concentration of B in moldm3
Mass Conc(8dm³) = molar Conc × molar mass
Molar Conc of B = mass conc/molar mass = 19.2gdm³/127gmol
Cb = 0.151moldm3
Concentration of A in moldm-³, CA is given by CAVA/CBVB = na/nb
Where CA = ?
VA = 15.30cm³,
na = 1
CB = 1.51moldm-³, VB = 25.0cm³, nb = 5
CA × 15.30/0.151 × 250= 1/5
Therefore CA = 0.151×25×1/15.30×5 = 3.775/76.50 = 0.049moldm-³
Hence Conc. Of A in moldm-³ = 0.049moldm-³
No of moles of Fe²+ in the volume of B pipetted
n= cv/1000 =
= 0.003775moles of Fe²+
TEST: C + about 5cm3 of distilled water
OBSERVATION: C dissolves to form a clear solution;
Turns blue litmus paper to pink
INFERENCE: C is a soluble salt;
C is fairly acidic
TEST: 1st portion of C + NaOH(aq) in drops;
OBSERVATION: pale-blue precipitate forms in drops of NaOH(aq) which persists in excess of NaoH(aq)
INFERENCE: cu²+ present.
TEST: 2nd portion of C + NH3(aq) in drops and then in excess
OBSERVATION: pale blue precipitate forms in drops of NH3(aq).
Ppt dissolves to form a deep blue coloration.
INFERENCE: Cu²+ present.
TEST: 3rd portion of c + AgNO3(aq) + HCL(aq)
OBSERVATION: a creamy white ppt forms.
INFERENCE: Halogen/Halides present.
TEST: D + about 5cm³ of distilled H2O
OBSERVATION: D dissolves to form a clear solution.
The test tube feels warmth
INFERENCE: D is a soluble salt.
TEST: 2cm³ of D + HCL(aq)
OBSERVATION: effervescence occurs, an odourless and colourless gas evolved. Precipitate dissolves in dil HCL(aq)
INFERENCE: CO32-, so32-present
A white precipitate of barium sulphate will be formed by the instant reaction between sulphuric acid and barium chloride.
H2SO4(aq) + BaCl2(aq) -> BASO4 + 2HCl
Iron II sulphide reacts with hydrochloric acid, releasing the malodorous and very toxic gas, hydrogen sulphide
FeS(aq) + 2HCl -> Fecl2(aq) + H2S(g)
When solid iron fillings are added to dilute aqueous hydrochloric acid, Iron(II)Chloride OR ferrous chloride is formed, with the liberation of hydrogen gas.
Fe(s) + 2HCL(aq) -> Fecl2 + H2(g)