Explore topic-wise InterviewSolutions in .

Answer:  Lenard

ANSWER : 13.712 g mol^-1

Given : 

Vapour density (VD) of metal chloride = 79

Metal in metal oxide = 33%

To Find : 

Atomic weight of metal = ? 

Solution : 

Let the weight of metal oxide be 100 g.

So the weight of metal will be 33 g and weight of oxygen will be 77.

So equivalent weight (E_w) of metal 

\(= {weight\:of\:metal \over weight\:of\: oxygen } × 8\)

\(={ 33\over77} × 8\)

= 3.428 g

We know that , 

Valency  \(= {2 ×VD\over {E}_{w}+35.5}\)

^{​​​​\( = {2×79 \over 3.428 + 35.5}\)}

^{\(= {158\over 38.928}\)}

^{= 4.05}

^{≈ 4}

^{Now ,}

^{Atomic weight = valency × E_w} 

= 4 × 3.428 

= 13.712 g mol^-1

Hence The atomic weight of metal is 13.712 gmol^-1

Correct Option (d) PH₃ < AsH₃ < NH₃ < SbH₃

Explanation:

Boiling points of nitrogen family generally increases on moving down the group but boiling point of NH₃ is higher than PH₃ and AsH₃ due to hydrogen bonding. Thus, order of boiling point is

 PH₃ < AsH₃ < NH₃ < SbH₃

(b) MnBr₄^2-, tetrahedral

Explanation:

In case of MnBr₄^2- the oxidation state of Mn is +2. Thus, probable formula and geometry is MnBr₄^2- and tetrahedral respectively (Br is a weak field ligand).

Correct option(1) 10000

Explanation:

Most probable velocity (Cp) = 0.8166×RMSvelocity

= 0.8166×12240 =10000 cm/ sec

q = it

q = 1 x 1

q =  1c

q = 1 e

n = q/2

n = \(\frac{1}{1.6\times10^{-19}}\) 

⇒ 0.625 x 10¹⁹ e/sec

(i) When the body is grey with ε = 0.42 :

T₁ = 1000 + 273 = 1273 K 

T₂ = 500 + 273 = 773 K 

ε at 1000°C = 0.42 

ε at 500°C = 0.72 

σ = 5.67 × 10^–8 

Heat loss per m² by radiation, 

q = εσ (T₁⁴ - T₂⁴) 

= 0.42 × 5.67 × 10^–8 [(1273)⁴ – 773)⁴] = 54893 W i.e., 

Heat loss per m² by radiation = 54.893 kW.

(ii) When the body is not grey : 

Absorptivity when source is at 500°C = Emissivity when body is at 500°C i.e., absorptivity, α = 0.72

Then, energy emitted = εσ T₁⁴ = 0.42 × 5.67 × 10^–8 × (1273)⁴

Energy absorbed = ασT₂⁴ = 0.72 × 5.67 × 10^–8 × (773)⁴ i.e., 

q = Energy emitted – Energy absorbed 

= 0.42 × 5.67 × 10^–8 × (1238)⁴ – 0.72 × 5.67 × 10^–8 × (773)⁴ 

= 62538 – 14576 = 47962 W 

i.e., Heat loss per m² by radiation = 47.962 kW.

(i) Judicious use of aerosol spray propellants such as fluro carbons and chloroflurocarbons which cause depletion or hole in ozone layer.

(ii) Control over large scale nuclear explosions and limited use of supersonic planes.

Skin cancer is caused in human beings due to the depletion of ozone layer in the atmosphere.

Aluminium being a reactive metal readily combines with oxygen to give a protective layer of Al₂O₃. As a result further corrosion stops.

There will be formation of holes in the copper vessel after several days. In terms of activity series, copper is more reactive than silver. Hence copper will react with silver nitrate to form copper(II) nitrate and silver metal. Thus, there will be deposition of silver. 

2AgNO₃(aq.) + Cu(s)→ Cu(NO₃)₂(aq.) + 2Ag(s).

(i) Though hydrogen is not a metal but even then it has been assigned a place in the activity series. The reason is that like metals, hydrogen also has a tendency to lose electron and forms a positive ion H⁺
The metals which lose electrons less readily than hydrogen are placed below it and the metals which lose electrons more readily than hydrogen are placed above it in the reactivity series of metals.
(ii) By displacement reaction silver can be shown to be chemically less reactive than copper or copper is more reactive than silver. If a piece of silver is immersed in a solution of copper sulphate, no reaction will take place because silver is less reactive than copper and will not displace copper from the copper sulphate solution.
                              CuSO₄(aq) + Ag(s) ------> No reaction

On the other hand, if a copper plate is placed in a solution of silver nitrate, copper will slowly displace silver from the solution and blue solution of copper nitrate is formed.
  2AgNO₃(aq) + Cu(s) -----> Cu(NO₃)₂(aq) + 2Ag(s)
       Colourless                           Blue
This shows that copper is more reactive than silver'

The metals which occur above hydrogen in the activity series liberate hydrogen from dilute acids. Since Na, Fe and Pb occur above hydrogen in the activity series, they will liberate hydrogen from dilute acids. Cu and Hg occur below hydrogen in the activity series and hence they will not liberate hydrogen from dilute acids.

Only metals form basic oxides. Out of the elements Si, S, Se, P and Mg, only Mg is a metal and other elements are either non-metals or semi-metals. Hence magnesium will form a basic oxide, MgO.

Correct answer is (c) Magnesium hydroxide

\(\underset{\text{Magnesium ribbon}}{2Mg} + \underset{\text{oxygen}}{O_2}\) \(\overset{\text{burn}}{\longrightarrow}\, \underset{\text{Magnesium oxide}}{2MgO}\)

due to, wet watch glass, MgO reacts with H₂O and form magnesium hydroxide -

\(MgO\, + H_2O\, \longrightarrow \, \underset {\text{Magnesium Hydroxide}}{Mg(OH)_2}\)

Correct answer is (a) A - no change; B - becomes dark red; C - becomes blue

In distilled water neighter red litmus nor blue litmus paper, changes their colours.

In acidic medium - Red litmus paper becomes dark red.

In basic medium - Red litmus paper - becomes blue.

Correct answer is (c) Dry the gas

2 Nacl + H₂SO₄ \(\overset{\triangle}{\longrightarrow}\) Na₂SO₄ + 2HCl↑

Calcium chloride is used to dry the HCl gas.

Calcium chloride is highly hygroscopic in nature.

Correct answer is (b) Near the cathode, at the anode, at the cathode

\(2NaCl \,(aq) + 2H_2O \,(l)\) \(\overset{\text{elecrolysis}}{\longrightarrow}\) \(\underset{(A)\\ \text{(near cathode)}}{2NaOH\,(aq)} + \underset{(B)\\ \text{(Anode)}}{Cl_2\,(g)} + \underset{(C)\\ \text{(cathode)}}{H_2\,(g)}\)

When electricity is passed through an aquious solution of sodium chloride (brine), chlorine gas is given off at the anode and Hydrogen gas at the cathode.

Sodium Hydroxide solution is formed near the cathode.

Correct answer is (a) I, II and IV

2 NaOH + H₂SO₄ \(\longrightarrow\) Na₂SO₄ + 2H₂O

(i) According to given chemical equation, we can clearly see that NaOH reacts with H₂SO₄ to produce Na₂SO₄ and water

(ii) The given chemical equation is a balance equation. we can clearly see that, one molecule of H₂SO₄, reacts with 2 molecule of NaOH and produced one molecule of Na₂SO₄ and two molecule of water (H₂O).

(iii) acids and bases are ionic in nature, in equations solution, they decomposed like that -

NaOH \(\overset{H_2O}{\longrightarrow}\) \(Na^{+}(aq) + \overset{\ominus}{OH}(aq)\)

\(H_2SO_4\, \overset{H_2O}{\longrightarrow}\,H_3O^+ +HSO_4{^-}{(aq)}\)

(iv) The given reaction is a example of double displacement reaction, also you can say that it is a neutralization reaction. There is no change in oxidation state of any atoms. that is why it is not a redox reaction.

Lead is the poorest electricity conducting metal and silver is the best electricity conducting metal.

(C) aluminium from Al₂O₃ 

(D) magnesium from MgCO₃.CaCO₃

Fe₂O₃ and SnO₂ undergoes C reduction