Purification and Organic Compounds Test

Result

Purification and Organic Compounds Test - 4

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Weekly Quiz Competition

Question 1
4 / -1

To an acidified dichromate solution, a pinch of Na₂O₂ is added and shaken. What is observed?

A
Blue colour

B
Red colour changing to green

C
Copious evolution of oxygen

D
Bluishgreen precipitate

Solution

Blue colour is due to the formation of CrO₅

Na₂Cr₂O₇ + 4Na₂O₂ + 5H₂SO₄ → 2CrO₅ + 5Na₂SO₄ + 5H₂O.
Question 2
4 / -1

A mixture of zinc oxide and cupric oxide can be separated by

A
boiling the mixture with caustic potash solution and filtering.

B
boiling the mixture with dilute HNO₃ and filtering.

C
heating with carbon

D
removing cupric oxide with magnet

Solution

ZnO is an amphoteric oxide while CuO is slightly basic oxide. When the mixture will be boiled with NaOH solution, ZnO will dissolve in it as sodium zincate solution while CuO will remain undissolved, which can be filtered off.

ZnO + 2NaOH →Na₂ZnO₂ + H₂O

water soluble

∴ (a)
Question 3
4 / -1

The effective component of bleaching powder is

A
Ca²⁺

B
Cl^–

C

D
OCl^–

Solution

Bleaching powder is in fact a mixture of Ca(OCl)₂ and CaCl₂.Ca(OH)₂.H₂O. The active constituent is CaOCl₂, which on treatment with acid liberates Cl₂.
Question 4
4 / -1

When the solution of an inorganic salt in sodium hydroxide is boiled, a pungent gas is produced, which turns moist red litmus paper blue and a mercurous nitrate solution black. This indicates the presence of ________ ion in the salt.

A

B

C

D

Solution

Since the gas evolved turns red litmus paper blue, this implies that the gas is basic and as the gas also turns mercury(I) nitrate solution black, thus, the gas is NH₃ and the ion of the salt is
Question 5
4 / -1

The constituents of which of the following pairs of ions in a dilute HCl medium cannot be separated by passing H₂S gas through it?

A

B
Cu²⁺ and Cd²⁺

C
Zn²⁺ and Sn²⁺

D
Co³⁺ and Cu²⁺

Solution

That pair of ions cannot be separated by passing H₂S gas in dilute HCl medium in which either both the ions belong to II group or both of them do not belong to II group. The option (b) has both ions (Cu²⁺and Cd²⁺) belonging to II group.
Question 6
4 / -1

Highly pure dilute solution of sodium in liquid ammonia

A
shows blue colour

B
exhibits no electrical conductivity

C
produces sodamide

D
produces hydrogen gas

Solution

Anhydrous liquid ammonia is analogous to water as a solvent. It is also associated through hydrogen bonding of course the NHN bond being somewhat weaker than OHO bond in water. The earth metals (except Be) dissolve in liquid ammonia, without chemical reaction, giving blue solutions, which are strongly reducing in character due to the following reaction.

Dilute metal ammonia solutions are blue in colour due to solvated electrons. All metal ammonia solution posses high electrical conductivity, which varies with the metal concentration. At large dilution the conductivity of these solution exceeds that of a fully ionized salt in water while in most concentrated solutions, the conductivity is almost the same as that for a pure metal. The dilute solutions are paramagnetic indicating presence of unpaired electrons in the solution.
Question 7
4 / -1

Heating (NH₄)2Cr₂O₇ yields a gas, which is also obtained by

A
Mg₃N₂ in H₂O

B
heating NH₄NO₂

C
heating NH₄NO₃

D
Na₂O₂ in H₂O

Solution

All ammonium salts liberate NH₃ on heating except NH₄NO₂, (NH₄)₂Cr₂O₇ and NH₄NO₃. Any metal nitride on treatment with water gives corresponding hydroxide and NH₃ gas.
Question 8
4 / -1

A sodium salt of an unknown anion when treated with MgCl₂ gives a white precipitate, only on boiling. The anion is

A

B

C

D

Solution

Bicarbonate, nitrate and sulphate of magnesium are water– soluble. Bicarbonates on heating dissociate to give carbonates, which are insoluble.
Question 9
4 / -1

[X] + H₂SO₄→ [Y] a colourless gas with suffocating smell

[Y] + K₂Cr₂O₇ + H₂SO₄ → green solution [X] and [Y] is

A

B
Cl^– , HCl

C
S^2– , H₂S

D

Solution

A colorless suffocating gas, which on passing through acidified K₂Cr₂O₇ solution, turns the solution green. Thus, the gas would be SO₂ and the salt must be (sulphite).
Question 10
4 / -1

A solution is 10^–3 M each in Mn²⁺, Fe²⁺, Zn²⁺ and Hg²⁺. It is treated with H₂S in alkaline medium. If K_sp of MnS, FeS, ZnS and HgS are 10^–15, 10^–23, 10^–20 and 10^–54 respectively, which one will precipitate first?

A
FeS

B
MgS

C
HgS

D
ZnS

Solution

The sulphide with the least solubility product would be precipitated first as all the metal ions in the solution possess same charge and have same concentration.