Tag: chemistry

Questions Related to chemistry

Which of the statement about aluminium is not correct?

  1. It liberates hydrogen from acid as well as alkalies

  2. It liberates hydrogen from acid but not from alkalies

  3. It liberates hydrogen from hot alkali solution

  4. It liberates hydrogen from boiling water


Correct Option: B
Explanation:

Solution:- (B) It liberates hydrogen from acid but not from alkalies

Aluminium is an amphoteric metal and hence it reacts with both acids as well as bases.
Reaction of $Al$ with acid-
$2 Al + HCl \longrightarrow 2 Al{Cl} _{3} + 3 {H} _{2}$
Reaction of $Al$ with base-
$2 Al + 2 NaOH + 2 {H} _{2}O \longrightarrow 2 NaAl{O} _{2} + 3 {H} _{2}$

The common impurities present in bauxite are:

  1. ${ Fe } _{ 2 }{ O } _{ 3 },{ SiO } _{ 2 }$

  2. $NaCl,{ MgCl } _{ 2 }$

  3. ${ AlCl } _{ 3 },{ MgCl } _{ 2 }$

  4. ${ CaCl } _{ 2 },{ MgCl } _{ 2 }$


Correct Option: A
Explanation:

Bauxite has a number of impurities in it including iron oxides(hematite and goethite), $Fe _2O _3$ the sand silicon dioxide $SiO _2$ the clay mineral Kaolinite and asmall amount of $TiO _2$ known as Anatase.

Hence option A is correct answer.

In the purification of $Al$ by Hoop's process the correct statement is 

  1. electrolytic cell is iron and it contains three layer mass

  2. upper layer is pure $Al$ and Carbon rods in it are cathode and the bottom layer contains impure $Al$ and carbon lining of the cell is anode

  3. Electrolyte is the middle layer with fused mixture of flouride.

  4. All of the above


Correct Option: D

In thermite welding, aluminium acts as a/an:

  1. solder

  2. flux

  3. oxidising agent

  4. reducing agent


Correct Option: D
Explanation:

In thermite welding, aluminium acts as reducing agent. Aluminium is used in thermite welding of broken iron parts. The reduction of ferric oxide by aluminium is highly exothermic and therefore, the iron formed will be in the molten state. Thermite reaction involving aluminium is also called Goldschmidt alumino thermic reaction.

Read the following statements and choose the correct alternative.
Statement I :  Aluminium fool cannot bebe used in $\alpha $-particle - particle scattering experiment.
Statement II : Aluminium is a highly malleable metal.

  1. Both statement are true and statement II is a correct reason of Statement I.

  2. Both statement are true and statement II is a not correct reason of Statement I.

  3. Statement I is true and statement II is false

  4. Statement II is true and statement I is false


Correct Option: C
Explanation:
  • If Aluminium foil was used the,the scattering angles would have changed, but the qualitative results would also change: the reason Rutherford chose gold was because it is EXTREMELY malleable. One can stretch gold foil until it is only a few atoms thick in places, which is not possible with aluminum.
  • Hence option C is correct answer.

${ aAl } _{ 2 }{ \left( { C } _{ 2 }{ O } _{ 4 } \right)  } _{ 3 }(s)\xrightarrow [  ]{ \Delta  } \quad { bAl } _{ 2 }{ O } _{ 3 }(s)+cCO(g)+d{ CO } _{ 2 }(g)$
According to the equation for the reaction represented above, what is the mole of $CO$ to ${CO} _{2}$ that is produced by the decomposition of aluminium oxalate?

  1. $1$ mole $CO$; $1$ mole of ${CO} _{2}$

  2. $1$ mole $CO$; $2$ moles of ${CO} _{2}$

  3. $1$ mole $CO$; $3$ moles of ${CO} _{2}$

  4. $2$ moles $CO$; $1$ mole of ${CO} _{2}$

  5. $3$ moles $CO$; $1$ mole of ${CO} _{2}$


Correct Option: A
Explanation:

${ Al } _{ 2 }{ ({ C } _{ 2 }{ O } _{ 4 }) } _{ 3 }(s)\xrightarrow { \Delta  } { Al } _{ 2 }{ O } _{ 3 }(s)+3CO(g)+3C{ O } _{ 2 }(g)$

According to the equation for the reaction represented above, what is the mole of $CO$ to $C{ O } _{ 2 }$ that is produced by the decomposition of aluminium oxalate is 1 mole of $C{ O }$ : 1 mole of $C{ O } _{ 2 }$

X$ _2$O$ _3$ is produced if ratio of element to oxygen is 2:3 in the reaction of element and oxygen,what is that element?

  1. Ca

  2. Li

  3. Al

  4. F

  5. C


Correct Option: C
Explanation:

$\displaystyle   X _2O _3$ is produced if ratio of element to oxygen is 2:3 in the reaction of element and oxygen, that element is Aluminum (Al). It forms $\displaystyle   Al _2O _3$ . The valence of Al (and X)  is +3 and that of oxygen is -2.
Ca, Li, F and C have valencies +2, +1, -1 and 4 respectively.

The dissolution of $Al(OH) _3$ by a solution of NaOH results in the formation of:

  1. $[Al(H _2O) _2(OH) _4]^-$

  2. $[Al(H _2O) _3(OH) _3]^+$

  3. $[Al(H _2O) _4(OH) _2]^+$

  4. $[Al(H _2O) _6(OH) _3]^-$


Correct Option: A
Explanation:

1. $[Al(H _2O) _4(OH) _2]^+$

2. $Al(OH) _3$  dissolves in $NaOH$ solution to give $Al(OH)^{-4}$ ion which is supposed to have the octahedral complex species $[Al(OH) _4(H _2O) _2]^-$ in aqueous solution.

3. $Al(OH) _3 + NaOH(aq) \longrightarrow  [Al(OH) _4(H _2O) _2]^-(aq) + NaOH^+(aq)$

Aluminium chloride exists as dimer, $Al _{2}Cl _{6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives:

  1. $[Al(OH) _{6}]^{3-} + 3HCl$

  2. $Al _{2}O _{3} + 6HCl$

  3. $Al^{3+} + 3Cl^{-}$

  4. $[Al(H _{2}O) _{6}]^{3+} + 3Cl^{-}$


Correct Option: D
Explanation:

When ${ Al } _{ 2 }{ Cl } _{ 6 }$ is dissolved in water, it is ionized to form ${ \left[ Al{ \left( { H } _{ 2 }O \right)  } _{ 6 } \right]  }^{ 3+ }$ and ${ Cl }^{ \left( - \right)  }$ due to large hydration energy of ${ Al }^{ 3+ }$.

${ Al } _{ 2 }{ Cl } _{ 6 }+12{ H } _{ 2 }O\rightleftharpoons 2{ \left[ Al{ \left( { H } _{ 2 }O \right)  } _{ 6 } \right]  }^{ 3+ }+{ 6Cl }^{ \left( - \right)  }$

Which of the following is soluble in water?

  1. $Na _2CO _3$

  2. $CaCO _3$

  3. $ZnCO _3$

  4. $Al _2(CO _3) _3$


Correct Option: A