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)
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