(1ai) 

Methyl Orange

Reason – Titration of strong acid against base

(1aii) 

Yellow

(1aiii)

Hydrochloric acid

(1aiv)2HCL(aq)+ x2Co3 -> 2 x CL(aq) + H20(aq) + CO2(g)

(1bi) Molar mass of HcL = 1+35.5

=36.5gmol^-1

Con of A is moldm^-3 =

Con of A of gdm^3/molar mass in gdm^3

=7.30gdm^-3/36.5gmol^-1

=0.200moldm^-3

(1bii)

CaVa/CbVb=na/nb

Ca=0.200mol/dm^3

Va=26.20cm^3

Cb=?

Vb=25.0cm^3

na=2

nb=1

(0.200*26.20)/(Cb*25.0)=2/1

Cb=(0.200*26.20*1)/(25.0*1)

Cb=5.24/36.5

Cb=0.2098mol/dm^3 

(1biii)

Conc. of B in gdm-3/Conc. of B in moldm-3

=10.6gdm-3/0.148 gdm-3

=101.15g

=101gmol-1

(1biv)

Relative atomic mass of X

2x+12+17.8=101

2x=101-29.8

2x=71.2

x=35.6

==================================

(2ai)

INFERENCE: Sample X is a soluble salt

(2aii)

OBSERVATION: A gas is evolved. The gas turns red moist litmus blue. 

It forms white fumed when NaOH solution was added was added

INFERENCE: Alkaline gas

(2aiii)

INFERENCE: CO3 gas is CO3^2-

(2aiv)

INFERENCE: The solution is alkaline

(2av)

(NH4)2CO3(aq)+2NaOH(aq) >> Na2CO3(aq)+2NH3+2H2O(g)

(2bi)

REAGENTS: Adition of KSCN Solution

(Fe2+): Pale-colouration

(Fe3+) Blood red colouration

(2bii)

REAGENT: Addition of NaOH solution in drop till excess

(Fe2+): Dirty green gelatineous precipitaye which is insolube in excess NaOH.

(Fe3+): Reddish brown gelatineous precipitate which is insolube in excess NaOH.

(2biii)

REAGENT: Addition of K2Fe(CN) solution

(Fe2+): It forms a dark blue precipitate

(Fe3+): A deep red colouration

====================================

(3a)

-NH4Cl and AgNo3

-ZnCo3

-FeS

-AgNO3

-NH4Cl

(3b)

-Burrette

-Retort stand

-Solution/Acid

-Conical flask

CHEMISTRY QUESTIONS

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