TN 12TH CHEMISTRY - Online Test

Q1. pH of a saturated solution of Ca(OH)2 is 9. The Solubility product ( Ksp )of Ca(OH)2
Answer : Option A
Explaination / Solution:

Ca(OH)2 Ca2+ + 2OH-

Given that pH = 9

pOH = 14-9 = 5

[pOH = -log10 [OH- ]]

[OH- ] = 10- pOH

[OH- ]=10-5 M

Ksp =[Ca2+ ][OH- ]2

= (10-5 /2 )×(10-5 )2

=0.5 ×10-15


Q2. The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76 ×103 S cm1 at 298 K is
Answer : Option B
Explaination / Solution:

Λ = κ/M ×103 mol1 m3

= [( 5.76 × 103 S cm1 ×103 ) / (0. 5)]  mol1m3

= [(5.76 × 103 × 103 ×106 ) / (0.5)] S cm1mol1 cm3 .

= 11.52 S cm2 mol1


Q3.

Assertion : Coagulation power of Al3+ is more than Na+ .

Reason : greater the valency of the flocculating ion added, greater is its power to cause precipitation
Answer : Option A
Explaination / Solution:

(Hardy-Schulze rule)

Q4. Which one of the following is the strongest acid
Answer : Option C
Explaination / Solution:
No Explaination.


Q5.
Answer : Option C
Explaination / Solution:

X-HCHO

Y-(CH2 )6N4


Q6.
Answer : Option C
Explaination / Solution:



Q7.

Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence of little hydrochloric acid, it becomes levorotatory.

Reason: Sucrose hydrolysis gives equal amounts of glucose and fructose. As a result of this change in sign of rotation is observed.

Answer : Option A
Explaination / Solution:
No Explaination.


Q8. Which one of the following structures represents nylon 6,6 polymer?
Answer : Option D
Explaination / Solution:
No Explaination.


Q9. Conjugate base for Bronsted acids H2O and HF are
Answer : Option C
Explaination / Solution:

H2O + H2O H3O+ + OH-

acid 1 + base 1 acid 2 + base 2


HF + H2O H3O+ + F-

acid 1 + base 1 acid 2 + base 2

Conjugate bases are OH- and F- respectively


Q10.
Calculate ΛºHOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25oC .
Answer : Option C
Explaination / Solution:

( Λº )HoAC = ( Λº)HCl + ( Λº)NaOAC - ( Λº)NaCl

= (426.2 + 91) (126.5)

= 390.7