Chemistry Test 141

Result

Chemistry Test 141

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Place the following alcohols in decreasing order of rate of dehydration with concentrated H₂SO₄ :

1. CH₃CH₂CH(OH)CH₂CH₂CH₃

2. (CH₃)₂C(OH)CH₂CH₂CH₃

3. (CH₃)₂C(OH)CH(CH₃)₂

4. CH₃CH₂CH(OH)CH(CH₃)₂

5. CH₃CH₂CH₂CH₂CH₂CH₂OH

A
3 > 2 > 4 > 5 > 1

B
3 > 2 > 4 > 1 > 5

C
3 > 2 > 1 > 4 > 5

D
3 > 2 > 1 > 5 > 4

Solution

The alcohols (3) and (2) are both 3^o, but alcohol (3) gives a more substituted alkene. Alcohol (4) and (1) are both 2^o, but alcohol (4) can give a more substituted alkene and alcohol (5) is 1^o. Rate of dehydration of alcohols with concentrated H₂SO₄ follows the order 3^o > 2^o > 1^o.
Question 2
4 / -1

1.2 g of a salt with its empirical formula K_xH_y(C₂O₄)_z was dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.

A
y = 3x

B
y = 2x

C
y = 4x

D
y = 5x

Solution
Question 3
4 / -1

The structure of H₂O₂ is

A
Spherical

B
Non - planar

C
Planar

D
Linear

Solution

In H₂O₂, the O - H groups are not in the same plane. So it has non - planar structure. If has a half - opened book structure in which the two O - H groups lie on the two pages of the book. The angle between two pages of the book is 94^o and H - O - O bond angle is 97^o.
Question 4
4 / -1

In nitroprusside ion the iron and NO exist as Fe II and NO⁺rather than Fe III and NO. These forms can be differentiated by

A
Estimating the concentration of iron

B
Measuring the concentration of CN^-

C
Measuring the solid state magnetic moment

D
Thermally decomposing the compound

Solution

Fe II and Fe III will have different values of magnetic moment due to different number of unpaired electrons in their d-orbitals.
Question 5
4 / -1

Among the following pair of oxides, which pair cannot be reduced by carbon to give the respective metals ?

A
Fe₂O₃, ZnO

B
PbO, Fe₃O₄

C
Cu₂O, SnO₂

D
CaO, K₂O

Solution

Potassium and calcium are strong reductant, hence their oxides cannot be reduced by carbon.
Question 6
4 / -1

Two liquids A and B are mixed. The partial vapour pressures of A and B in pure state are 100 and 200 mm respectively. If they are mixed in 1 : 4 mole ratio, assuming that mixture obeys Raoult's law, the mole fractions of A and B present in gaseous state in equilibrium of above solution are :

A

B

C

D

Solution
Question 7
4 / -1

Identify the final product (Z) in the following sequence of reactions :

A
C₆H₅CH₂NH₂

B
C₆H₅CH₂CH₂NH₂

C
C₆H₅CH₂CH₂CH₂NH₂

D

Solution
Question 8
4 / -1

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

A
A₂B

B
AB₂

C
A₂B₃

D
A₂B₅

Solution

In FCC arrangement
after removal of atom from face centre.
No. of A atoms per unit cell = 1/8 × 8 = 1
No. of B atoms per uni cell = = 1/2 ×5= 5/2
so formula is A₂B₅
Question 9
4 / -1

The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25. The equilibrium constant of the reaction is-
(F = 96500 C mol^-1 ,R = 8.314 JK^-1 mol^-1

A
1.0 × 10¹

B
1.0 × 10⁵

C
1.0 × 10¹⁰

D
1.0 × 10³⁰

Solution

∆G° = ∆G + RT lnQ
-nFE° = -nFE + RT lnQ
E_Cell = E°_Cell - RT/nF lnQ
At equilibrium, E = 0
0 = 0.591 - (0.0591/1)× logK_C
-0.591 = -0.0591 logK_C
log K_C = 0.591/0.0591 = 10
K_C = antilog 10 = 1×10¹⁰
Question 10
4 / -1

For the indicator HIn the ratio [In_-]/[HIn] is 7.0 at pH of 4.3. K_eq for the indicator is [Given log 7 = 0.845 and Antilogo (0.545) = 3.5

A
3.5 × 10^-4

B
3.5 × 10^-5

C
3.5 × 10^-2

D
3.5 × 10^-3

Solution