# Q&A - Ask Doubts and Get Answers

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Medical
140 Views   |

Which one of the following is present as an active ingredient in bleaching powder for bleaching action?

• Option 1)

$C_{a}OCl_{2}$

• Option 2)

$C_{a}\left ( OCl \right )_{2}$

• Option 3)

$C_{a}O_{2}Cl$

• Option 4)

$C_{a}Cl_{2}$

2

Medical
111 Views   |

Which of the following compounds has the lowest melting point?

• Option 1)

$C_{a}Cl_{2}$

• Option 2)

$C_{a}Br_{2}$

• Option 3)

$C_{a}I_{2}$

• Option 4)

$C_{a}F_{2}$

As we learnt in Nature of compounds formed by alkaline earth metals - Be and Mg form covalent compounds other elements froms ionic compounds. This character increases down the group. -    As the covalent character increases down the group the melting point among the given species.   Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is...
Medical
196 Views   |

Of the following, which one is classified as polyester polymer?

• Option 1)

Terylene

• Option 2)

Bakelite

• Option 3)

Melamine

• Option 4)

Nylone-66

As we learnt in  Terylene / Dacron - - Manufactured by heating ethylene glycol and terephthalic acid in the presence of zinc acetate antimony trioxide. - Step growth, Copolymer - wherein - Crease resistant  - Used in blending with cotton and wool fibres and in making helmets.    ethylene is classified as a polyester polymer because it is manufactured heating ethylene glycol and...
Medical
123 Views   |

Match List- I with List- II for the compositions of substances and select the correct answer using the code given below the lists:

List- I                                                         List- II

Substances                                               Composition

(1) plaster of paris                                     (i) $C_{a}SO_{4}.2H_{2}O$

(2) Epsomite                                              (ii) $C_{a}SO_{4}.\frac{1}{2}H_{2}O$

(3) Kieserite                                                (iii) $M_{a}SO_{4}.7H_{2}O$

(4) Gypsum                                                (iv) $M_{g}SO_{4}.H_{2}O$

(v) $C_{a}SO_{4}$

• Option 1)

Code :

(1)         (2)        (3)           (4)

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

• Option 2)

Code :

(1)         (2)        (3)           (4)

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

• Option 3)

Code :

(1)         (2)        (3)           (4)

(i)           (ii)       (iii)             (v)

• Option 4)

Code :

(1)         (2)        (3)           (4)

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

As we learnt in concept Plaster of paris - It is hemihydrate of calcium sulphate On mixing with water, it takes up the water of crystallisation and is converted into dihydrate which sets into hard mass - wherein formula:    plaster of paris: Epsomite: Kieserite: Gypsum: Option 1) Code : (1)         (2)        (3)           (4) (iii)          (iv)       (i)          (ii) This solution is...
Medical
218 Views   |

A solid compound XY has NaCI structure. If the radius of the cation is 100 pm, the radius of the anion $\left ( Y^{-} \right )$ will be:

• Option 1)

275.1 pm

• Option 2)

322.5 pm

• Option 3)

241.5 pm

• Option 4)

165.7 pm

As learnt in Relation between radius of constituent particle, R and edge length, a for body centered cubic unit cell - -    In a BCC lattice, the oppositely charged ion occupies the centre while the corners are occupied by other ions. So, the distance between the centre and the corner,                                                                                                              ...
Medical
94 Views   |

A gaseous mixture was prepared by taking equal mole of $CO_{2}\ and\ N_{2}$ . If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen $\left ( N_{2} \right )$ in the mixture is:

• Option 1)

0.5 atm

• Option 2)

0.8 atm

• Option 3)

0.9 atm

• Option 4)

1 atm

Partial Pressure of a gas in gaseous mixture - Partial Pressure = (Total Pressure of Mixture)  (mole fraction of gas)   -     Option 1) 0.5 atm This solution is correct  Option 2) 0.8 atm This solution is incorrect  Option 3) 0.9 atm This solution is incorrect  Option 4) 1 atm This solution is incorrect
Medical
111 Views   |

A bubble of air is underwater at temperature $15^{o}C$ and the pressure 1.5bar. If the bubble rises to the surface where the temperature is $25^{o}C$ and the pressure is 1.0 bar, What will happen to the volume of the bubble?

• Option 1)

Volume will become greater by a factor of 1.6

• Option 2)

Volume will become greater by a factor of 1.1

• Option 3)

Volume will become smaller by a factor of 0.70

• Option 4)

Volume will become greater by a factor of 2.5

Ideal Gas Law - - wherein P - Pressure V - Volume n - No. of Moles R - Gas Constant T - Temperature    We know that from ideal gas equation given K= 288K, = 293 K, Option 1) Volume will become greater by a factor of 1.6 This is correct answer Option 2) Volume will become greater by a factor of 1.1 This is incorrect answer Option 3) Volume will become smaller by a factor of...
Medical
196 Views   |

The van's Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively:

• Option 1)

Less than one and greater than one

• Option 2)

Less than one and less than one.

• Option 3)

Greater than one and less than one.

• Option 4)

Greater than one and greater than one.

As learnt Vant Hoff factor for dissociation - number of particles associated degree of association e.g. Carboxylic acid dimerized in non polar solvents like Hexane.   - wherein n = 2 for dimerization n = 3 for trimerization    AND   Vant Hoff factor for dissociation - Where is the no. of dissociated particles degree of association   - wherein     Van't Hoff factor for association, ...
Medical
674 Views   |

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 second respectively. The molecular mass of A is 49u. Molecular mass of B will be

• Option 1)

50.00u

• Option 2)

12.25u

• Option 3)

6.50u

• Option 4)

25.00u

Graham’s law of diffusion -   where r is rate of diffusion of gas , M is molar mass. -    As we know => Option 1) 50.00u This is incorrect answer Option 2) 12.25u This is correct answer Option 3) 6.50u This is incorrect answer Option 4) 25.00u This is incorrect answer
Medical
252 Views   |

The freezing point depression constant for water is -1.86°C m-1. If 5.00 g Na2SO4 is dissolved in 45.0g H2O, the freezing point is changed by -3.82°C. Calculate the van't Hoff factor for Na2SO4.

• Option 1)

2.05

• Option 2)

2.63

• Option 3)

3.11

• Option 4)

0.381

As we learned in concept Application of Vant Hoff factor - -     m =   Option 1) 2.05 This option is incorrect. Option 2) 2.63 This option is correct. Option 3) 3.11 This option is incorrect. Option 4) 0.381 This option is incorrect.
Medical
86 Views   |

Mole fraction of the solute in a 1.00 molal aqueous solution is:

• Option 1)

0.1770

• Option 2)

0.0177

• Option 3)

0.0344

• Option 4)

1.7700

As we learned in concept Mole Fraction - -    Mole fraction = Let's assume, we have 1 mole of solute Then, wt. of solvent = 1 kg Moles of solvent = Mole fraction of solute =                                      =                                      = 0.0177   Option 1) 0.1770 This option is incorrect. Option 2) 0.0177 This option is incorrect. Option 3) 0.0344 This option is...
Medical
535 Views   |

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m , the freezing point of the solution will be:

• Option 1)

-0.18°C

• Option 2)

-0.54°C

• Option 3)

-0.36°C

• Option 4)

-0.24°C3

As we learned in concept Application of Vant Hoff factor - -                         =   Option 1) -0.18°C This option is incorrect. Option 2) -0.54°C This option is incorrect. Option 3) -0.36°C This option is incorrect. Option 4) -0.24°C3 This option is correct.
Medical
498 Views   |

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 2.57x10-3 bar. The molar mass of protein will be (R=0.083 L bar mol-1 K-1)

• Option 1)

51022 g mol-1

• Option 2)

122044 g mol-1

• Option 3)

31011 g mol-1

• Option 4)

61038 g mol-1

As we learned in concept Mathematical Expression for Osmotic Pressure -   - wherein Representation of osmotic pressure.    Osmotic pressure, C = M =     Option 1) 51022 g mol-1 This option is incorrect. Option 2) 122044 g mol-1 This option is incorrect. Option 3) 31011 g mol-1 This option is incorrect. Option 4) 61038 g mol-1 This option is correct.
Medical
89 Views   |

Name the type the structure of silicate in which one oxygen atom of $\left [ SiO_{4} \right ]^4^-$  is shared?

• Option 1)

Linear chain silicate

• Option 2)

Sheet silicate

• Option 3)

Pyrosilicate

• Option 4)

Three dimensional

Pyrosilicates - Two tetrahedral units share one oxygen atom between them containing basic unit of anion - wherein E.g. Thortveitite    In pyrosilicity, two tetrahedral units share one oxygen atom between them containing basic unit of anion   Option 1) Linear chain silicate This is incorrect option Option 2) Sheet silicate This is incorrect option Option 3) Pyrosilicate This is correct...
Medical
85 Views   |

If the enthalpy change for the transition of liquid water to steam is $30 kJ mol^{-1} \ at \ 27^{o}C$, the entropy change for the process would be

• Option 1)

10 $J mol^{-1}K^{-1}$

• Option 2)

1.0 $J mol^{-1}K^{-1}$

• Option 3)

0.1 $J mol^{-1}K^{-1}$

• Option 4)

100 $J mol^{-1}K^{-1}$

Entropy for phase transition at constant pressure - - wherein Transition Fusion, Vaporisition, Sublimation Enthalpy Internal Energy Transitional temperature    Entropy change =   = 100     Option 1) 10 Incorrect Option 2) 1.0 Incorrect Option 3) 0.1 Incorrect Option 4) 100 Correct
Medical
93 Views   |

When 1 kg of ice at 0o C melts to water at  0o C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/C, is:

• Option 1)

$273\: cal/ K$

• Option 2)

$8 \times 104\: cal/K$

• Option 3)

$80 \:cal/K$

• Option 4)

$293\:cal/K$

As learnt in Entropy for solid and liquid - When heat is given to a substance to change its state at constant temperature. - wherein T in kelvin                           Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option is correct
Medical
120 Views   |

During an isothermal expansion, a confined ideal gas does $-150 J$ of work against its surroundings. This implies that:

• Option 1)

150 J heat has been removed from the gas

• Option 2)

300 J of heat has been added to the gas

• Option 3)

No heat is transferred because the process is isothermal

• Option 4)

150 J of heat has been added to the gas

Using First law in Isothermal Process -   - wherein Isothermal process Heat supplied is equal to work done.    Negative work done by gas implies work is done on the system. For isothermal process,  Hence, heat is given to the system. Option 1) 150 J heat has been removed from the gas This option is incorrect Option 2) 300 J of heat has been added to the gas This option is...
Medical
109 Views   |

The total number of atomic orbitals in fourth energy level of an atom is:

• Option 1)

8

• Option 2)

16

• Option 3)

32

• Option 4)

4

As learnt in Azimuthal Quantum Number(l) - For a given value of n, l can have n values ranging from 0 to n – 1, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ....( n –1) -     for n = 4, l = 0, 1, 2, 3  we have s, p, d & f orbitals Total orbitals =                      =  Option 1) 8 This solution is incorrect Option 2) 16 This solution is correct Option...
Medical
74 Views   |

A mass of diatomic gas $(\gamma =1.4)$ at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $27^{o}C$ to $927^{o}C$. The pressure of the gas in final state is:

• Option 1)

28 atm

• Option 2)

68.7 atm

• Option 3)

256 atm

• Option 4)

8 atm

As learnt in Equation of state -   - wherein On solving    In adiabatic process,  Option 1) 28 atm       This option is incorrect Option 2) 68.7 atm This option is incorrect Option 3) 256 atm This option is correct Option 4) 8 atm This option is incorrect
Medical
156 Views   |

If n=6, the correct sequence for filling of electrons will be:

• Option 1)

$ns \rightarrow (n-2)f\rightarrow (n-1)d \rightarrow np$

• Option 2)

$ns \rightarrow (n-1)d\rightarrow (n-2)f \rightarrow np$

• Option 3)

$ns \rightarrow (n-2)f\rightarrow np \rightarrow (n-1)d$

• Option 4)

$ns \rightarrow np (n-1)d \rightarrow (n-2)f$

As we learnt in  Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein   In the ground state of atoms, the orbitals are filled in order of their increasing energies. Clearly for n=6, we have the order 6s, 4f, 5d, 6p or ns, (n-2)f, (n-1)d, np Option 1) This solution is correct Option 2) This solution is incorrect Option...
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