# Today Waec 2020 Mathematics Answers

##### By Olive - August 19, 2020 - [ Waec ] 3ai)

Zinctrioxonitrate (v) – Zn(NO3)2

(3aii)

2Zn(NO3)2(g) –> 2ZnO(s) + 4NO2(g) + O2(g)

(3aiii)

The residue when it has yellow colour which will turn white on cooling

(3b)

Given; M1 = 1.0mol/dm³

V1 = ?

M2 = 0.2mol/dm³

V2 = 250cm³

Using M1V1 = M2V2

1 × V1 = 0.2×250

V1 = 50cm³

Procedure: Measure out 50cm³ of the stock solution, dilute it to 0.2mol/dm³ by adding 200cm³ of water.

(3c)

Al(SO4)3 will turn blue litmus paper (1a)

Burette reading (cm)³|1st reading|2na Reading|3rd reading|

Final |15.25|30.53|45.79|

Initial |0.00|15.25|30.53|

Volume of acids used |15.25|15.28|15.26

Average volume of acid used =15.25+15.26/2

=15.255cm³

=15.26cm³

Or =15.27cm³

(1b)

Given; mass con of A =5g/500cm³ = 5g/0.5dn³

CA=10g/dm³

A is HNO3

Therefore molar mass =1+14+(16*3)=15+48=63g/mol¹

Molarity of A = gram con/molar mass

CA=10/63=0.1587mol/dm3

(1bii)

Using CAVA/CBVB =Na/NB

With reacting equation ; HNO³ + NaOH—NaNO3+H2O

nA=1 nB=1

0.158715.26/CB25.00=1/1

25CB=0.1587*15.26

CB=0.1587*15.26/25

CB=0.09687mol/dm³

(1biii)

B is NaOH

Molar mass

23+16+1=40g/mol

Conc of B in g/dm³=molarity * molar mass

= 0.09687*40

=3.8748g/dm³

(1biv)

No of moles present in 250cm³ of NaOH is =molar conc * volume

=0.09687*250/1000

=0.0242moles

Mole ratio of NaOH and NaNO3 is 1;1

No; of mole of NaNO3 which reacted is 0.0242

Mass of NaNO3 formed =molar mass * no; of moles

=85*0.0242

=2.05grams

2a)

TEST: C+burning splint

OBSERVATION: Sample C burst into flame .It burns with non smoky blue flame without soot. Colorless gas that turns wet blue litmus paper faint red and turns like water milky is present.

INFERENCE: C is volatile and flammable. The gas is CO2 from combustion of saturated organic compound.

(2bi)

TEST: C + distilled water + shake

OBSERVATION: Clear or colorless solution is observed

INFERENCE: C is miscible with water

(2bii)

TEST: C + Acidified K2Cr207

OBSERVATION: Orange color of K2Cr207 solution turns pale green and eventually pale blue on cooling

INFERENCE: C is a reducing agent

(2d)

Compound belongs to the class of secondary alkanol

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(1a)

In a tabular form

Burette Reading|1st reading|2nd reading|3rd reading

Final |15.25|0.00|45.79|

Initial|0.00||15.25|30.53|

Volume of acid used |15.25|15.28|15.26

Average volume of acid used = 15.25 + 15.26/2

= 15.255cm³

=15.26cm³

(1bi)

Given: Mass of conc of A = 5g/500cm³ = 5g/0.5dm³

Ca = 10g/dm³

A is HNO3

Therefore; Molar mass = 1+14+(16×3)

= 15+48

=63g/mol

Molarity of A = gram conc/molar mass

Ca = 10/63 = 0.1587mol/dm³

(ii) Using CaVa/CbVb = nA/nB

With reacting equation:

HNO3 +NaOH–>NaNO3 + H2O

nA = 1, nB = 1

0.1587×15.26/Cb×25.00 = 1/1

25Cb = 0.1587×15.26

CB = 0.1587×15.26/25

CB = 0.09687mol/dm³