On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane= 100 g mol^-1 and of octane = 114 g mol^-1)

72.0 kPa
36.1 kPa
96.2 kPa
144.5 kPa

Solution

Glycerol on oxidation with bismuth nitrate produces

oxalic acid
glyceric acid
glyoxalic acid
meso oxalic acid

Solution

The biggest particulate matter is

HNO₃ droplets
Soot
H₄SO₂ droplets
Fly ash

If ? is the degree of dissociation of Na₂SO₄ , the Vant Hoff’s factor (i) used for calculating the molecular mass is

1 + ?
1– ?
1 + 2?
1 – 2?

Solution

Which one of the following processes will produce hard water ?

Saturation of water with MgCO₃
Saturation of water with CaSO₄
Addition of Na₂SO₄ to water
Saturation of water with CaCO₃

Solution

Permanent hardness of water is due to chlorides and sulphates of calcium and magnesium i.e CaCl₂, CaSO₄,MgCl₂ and MgSO₄.

Which among the following statements are correct with respect to adsorption of gases on a solid?
(i)The extent of adsorption is equal to kpⁿ according to Freundlich isotherm
(ii)The extent of adsorption is equal to to kp^1/n according to Freundlich isotherm
(iii)The extent of adsorption is equal to (1+bp)/ap according to Langmuir isotherm
(iv)At low pressure x/m is directly proportional to the pressure

(i) and (iii)
(i) and (iv)
(ii) and (iii)
(ii) and (iv)

Solution

200 mL of an aqueous solution of a protein contains its 1.26 g.The osmotic pressure of this solution at 300 K is found to be 2.57 × 10^-3 bar. The molar mass of protein will be (R = 0.083 L bar mol^-1 K^-1)

51022 g mol^-1
122044 g mol^-1
31011 g mol^-1
61038 g mol^-1

Solution

Nu faces steric hinderance on attacking the leaving group
electrons are delocalised in the benzene ring
reaction is thermodynamically controlled
C – X bond possesses a double bond character

Solution

Lone pair present at X can enter in the ring. This gives rise to double bond character in C – X bond attached to the ring .

The structure of diborane (B₂H₆) contains

four 2c-2e bonds and four 3c-2e bonds
two 2c-2e bonds and two 3c-3e bonds
two 2c-2e bonds and four 3c-2e bonds
four 2c-2e bonds and two 3c-2e bonds

Solution

A first order reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completed

5 hours
7.5 hours
10 hours
20 hours

Solution

During dehydration of alcohols to alkenes by heating with conc. H₂SO₄ the initiation step is

formation of carbocation
elimination of water
formation of an ester
protonation of alcohol molecule

Solution

The compound which gives off oxygen on moderate heating is :

cupric oxide
mercuric oxide
zinc oxide
aluminium oxide

Solution

Oxygen can be prepared by heating oxides of Hg, Pb,Ag, Mn and Ba.

2HgO \(\overset{\Delta }{\rightarrow}\) ₂2Hg + O₂

Sodium formate on heating yields

oxalic acid and H₂
sodium oxalate and H₂
CO₂ and NaOH
sodium oxalate

Solution

The rate equation for the reaction 2A + B ⟶ C is found to be: rate = k[A][B]. The correct statement in relation to this reaction is that the

rate of formation of C is twice the rate of disappearance of A
t_1/2 is a constant
unit of k must be s^–1
value of k is independent of the initial concentrations of A and B

Solution

The velocity constant depends on temperature only. It is independent of concentration of reactants.

R – NH – Br
H– CO – NBr₂
R– N = C =O
all of these

Solution

As compared to covalent compounds, electrovalent compounds generally have

Low melting points and low boiling points
Low melting points and high boiling points
High melting points and low boiling points
High melting points and high boiling points

Solution

Ionic compounds generally have high melting and boiling points because of the strong electrostatic force of attraction between oppositely charged ions.Consequently, a considerable amount of energy is required to overcome strong attractive.

Which of the following isomers will have the highest octane number?

CH₃– (CH₂)₆– CH₃

Solution

More branched is the alkane, higher will be its octane number because more branched chains have higher antiknocting tendency than the corresponding straight chain hydrocarbons. Among the given options 2,2,4-trimethyl pentane, will have the highest octane number,100.

s-electrons of the valence shell of some elements show reluctance in bond formation. Such elements are — and belong to — :

lighter,s-block
heavier,d-block
heavier,f-block
heavier,p-block

Phenol can be converted to o-hydroxy-benzaldehyde by

Kolbe’s reaction
Reimer-Tiemann reaction
Wurtz reaction
Cannizaro reaction

Solution

Read the passage given below and answer the question that follow :
The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule which provides us the following very important information about the molecule.
(I)Stability of molecule : The molecule is stable if number of bonding molecular orbital electrons (N_b ) is greater than the number of anti bonding molecular orbital electrons(N_a).
(II)Bond order = 12 ( N_b – N_a )A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
(III)Nature of the bond : Bond order 1, 2 and 3 corresponds to single, double and triple bonds respectively.
(IV)Bond length : Bond length decreases as bond order increases.
(V)Magnetic nature : If molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.

N₂ has greater dissociation energy than \(N_{2}^{+}\), whereas O₂ has a lower dissociation energy than \(O_{2}^{+}\) it is because –g

bond order is reduced when O₂ is ionized to \(O_{2}^{+}\) and bond order is increased when N₂ is ionized to \(N_{2}^{+}\)
bond order is increased when O₂ is ionized to \(O_{2}^{+}\) and bond order is decreased when N₂ is ionized to \(N_{2}^{+}\)
bond order is decreased when O₂ and N₂ are ionized to \(O_{2}^{+}\) and \(N_{2}^{+}\) respectively
None of these

Solution

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