Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

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 x 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

  • 31011 g mol-1


D.

31011 g mol-1

straight pi space equals space CRT

equals space fraction numerator straight w space straight x space 1000 over denominator straight M space straight x space straight V space left parenthesis in space mL right parenthesis end fraction space straight x space RT

open square brackets therefore space space straight C space equals space fraction numerator straight n space straight x space 1000 over denominator straight V space in space mL end fraction and space straight n space equals space straight w over straight M close square brackets

equals space fraction numerator 1.26 space straight x space 1000 space straight x space 0.083 space straight x space 300 over denominator 2.57 space straight x space 10 to the power of negative 3 end exponent space straight x space 200 end fraction

equals space 61038 space straight g space mol to the power of negative 1 end exponent
458 Views

12.

In qualitative analysis, the metals of group I can be separted from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl- concentration is 0.10 M. What will be the concentration of Ag+ and Pb2+ be at equilibrium? (Ksp for AgCl = 1.8 x 10-10, Ksp for PbCl2  = 1.7 x 10-5)

  • [Ag+]  = 1.8 x 10-7 M; [Pb2+] = 1.7 x 10-6 M

  • [Ag+]  = 1.8 x 10-11 M; [Pb2+] = 8.5 x 10-5 M

  • [Ag+]  = 1.8 x 10-9 M; [Pb2+] = 1.7 x 10-3 M

  • [Ag+]  = 1.8 x 10-9 M; [Pb2+] = 1.7 x 10-3 M


C.

[Ag+]  = 1.8 x 10-9 M; [Pb2+] = 1.7 x 10-3 M

straight K subscript sp space for space AgCl space equals left square bracket Ag to the power of plus right square bracket left square bracket Cl to the power of minus right square bracket

therefore space left square bracket Ag to the power of plus right square bracket space equals space fraction numerator 1.8 space straight x space 10 to the power of negative 10 end exponent over denominator 10 to the power of negative 1 end exponent end fraction

space equals space 1.8 space straight x space 10 to the power of negative 9 end exponent space straight M.

straight K subscript sp space for space PbCl subscript 2 space equals space left square bracket Pb to the power of 2 plus end exponent right square bracket left square bracket Cl to the power of minus right square bracket squared

therefore space left square bracket Pb to the power of 2 plus end exponent right square bracket space equals space fraction numerator 1.7 space straight x space 10 to the power of negative 5 end exponent over denominator 10 to the power of negative 1 end exponent space straight x space 10 to the power of negative 1 end exponent end fraction
space
equals space 1.7 space straight x space 10 to the power of negative 3 end exponent space straight Mstraight K subscript sp space for space AgCl space equals left square bracket Ag to the power of plus right square bracket left square bracket Cl to the power of minus right square bracket

therefore space left square bracket Ag to the power of plus right square bracket space equals space fraction numerator 1.8 space straight x space 10 to the power of negative 10 end exponent over denominator 10 to the power of negative 1 end exponent end fraction

space equals space 1.8 space straight x space 10 to the power of negative 9 end exponent space straight M.

straight K subscript sp space for space PbCl subscript 2 space equals space left square bracket Pb to the power of 2 plus end exponent right square bracket left square bracket Cl to the power of minus right square bracket squared

therefore space left square bracket Pb to the power of 2 plus end exponent right square bracket space equals space fraction numerator 1.7 space straight x space 10 to the power of negative 5 end exponent over denominator 10 to the power of negative 1 end exponent space straight x space 10 to the power of negative 1 end exponent end fraction
space
equals space 1.7 space straight x space 10 to the power of negative 3 end exponent space straight M
382 Views

13.

The pair of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?

  • straight O subscript 2 superscript minus comma space straight O subscript 2 superscript 2 minus end superscript space minus space Both space diamagnetic
  • straight O to the power of plus comma space straight O subscript 2 superscript 2 minus end superscript space minus space Both space Paramagnetic
  • straight O subscript 2 superscript plus comma space straight O subscript 2 superscript space space minus space Both space paramagnetic
  • straight O subscript 2 superscript plus comma space straight O subscript 2 superscript space space minus space Both space paramagnetic

C.

straight O subscript 2 superscript plus comma space straight O subscript 2 superscript space space minus space Both space paramagnetic

As O2+, O2, O2-, O and O+ have unpaired electrons, hence are paramagnetic. The species which have unpaired electrons has paramagnetic behaviour.

374 Views

14.

A solution contains Fe2+ Fe3+, and I- ions. This solution was treated with iodine at 35o C.Eo for Fe3+/ Fe2+ is +0.77 V and Eo for I2/ 2 I- = 0.536 V. The favourable redox reaction is 

  • I2 will be reduced to I-

  • There will be No redox reaction

  • I- will be oxidised to I2

  • I- will be oxidised to I2


C.

I- will be oxidised to I2

 2 space straight I to the power of minus space rightwards arrow space straight I subscript 2 space plus space 2 straight e to the power of minus space left parenthesis oxidation space half space minus space reaction right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight E subscript oxidation superscript straight o space equals space minus 0.536 space straight V

Fe to the power of 3 plus end exponent space plus space straight e to the power of minus space space rightwards arrow space Fe to the power of 2 plus end exponent space left parenthesis reduction space half space minus space reaction right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight E subscript reduction superscript straight o space equals negative space 0.77 space straight V
______________________________________
space 2 Fe to the power of 3 plus end exponent space plus space 2 straight I to the power of minus space rightwards arrow space 2 space Fe to the power of 2 plus end exponent space plus space straight I subscript 2 semicolon
straight E to the power of straight o space equals space straight E subscript oxidation superscript straight o space plus space straight E subscript reduction superscript straight o
plus space ve
So comma space reaction space will space take space place.


plus space 2 space straight I to the power of minus space rightwards arrow space

620 Views

15.

The rate of reaction 2N2O3 → 4 NO2 + O2 can be written in three ways

fraction numerator negative straight d left square bracket straight N subscript 2 straight O subscript 5 right square bracket over denominator dt end fraction space equals space straight k space left square bracket straight N subscript 2 straight O subscript 5 right square bracket

fraction numerator straight d left square bracket NO subscript 2 superscript 7 right square bracket over denominator dt end fraction space equals space k apostrophe left square bracket N subscript 2 O subscript 5 right square bracket

fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space straight k apostrophe apostrophe space left square bracket straight N subscript 2 straight O subscript 5 right square bracket

The relationship between k and k' and between k and k'' are 

  • k' = 2k ; k' = k

  • k' = 2k; k" = k/2

  • k' = 2k; k' = 2k

  • k' = 2k; k' = 2k


B.

k' = 2k; k" = k/2

Rate space equals space minus 1 half fraction numerator straight d left square bracket straight N subscript 2 straight O subscript 5 right square bracket over denominator dt end fraction space equals space 1 fourth fraction numerator straight d left square bracket NO subscript 2 right square bracket over denominator dt end fraction space equals space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction

rightwards double arrow space 1 half straight k left square bracket straight N subscript 2 straight O subscript 5 right square bracket space equals space 1 fourth straight k apostrophe left square bracket straight N subscript 2 straight O subscript 5 right square bracket space equals space straight k " space left square bracket straight N subscript 2 straight O subscript 5 right square bracket

rightwards double arrow space straight k over 2 space equals space fraction numerator straight k apostrophe over denominator 2 end fraction space equals space straight k "

straight k apostrophe space equals space 2 space straight k semicolon space straight k " space equals space straight k over 2
372 Views

16.

The unit of rate constant for a zero order reaction is 

  •  mol L-1 s-1

  • L mol-1 s-1

  • L2 mol-2 s-1

  • L2 mol-2 s-1


B.

L mol-1 s-1

For zero order reaction,
Rate = k [ reaction]o
therefore, Rate = k
and unit of k = mol L- s-1

300 Views

17.

Which of the following carbonyl will have the strongest C- O bond? 

  • (Mn(CO)6+

  • Cr(CO)6

  • V (CO)6-

  • V (CO)6-


A.

(Mn(CO)6+

As the positive charge on the central metal atom increase, the less readily the metal can donate electron density into the anti - bonding pi-orbitals of CO ligand to weaken the C-O bond. Hence, the C-O bond would be strongest in Mn(CO)6+.

1472 Views

18.

The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.

  • Fe2O3 (s) + 3CO (g) → 2 Fe (l) +3CO2 (g)

  • CaCO3 (s) → CaO (s) + CO2 (g)

  • CaO (s) +SiO2 → CaSiO3 (s)

  • CaO (s) +SiO2 → CaSiO3 (s)


C.

CaO (s) +SiO2 → CaSiO3 (s)

A slag is an easily fusible material which is formed when gangue still present in the roasted or the calcined ore combines  with the flux. For example, in the metallurgy of iron, CaO (flux) combines with silica gangue to form easily fusible calcium silicate (CaSiO3) slag.

CaO + SiO2 → CaSiO3 (slag)

434 Views

19.

Which of the following oxide is amphoteric?

  • SnO2

  • CaO2

  • SiO2

  • SiO2


A.

SnO2

SnO2 is an amphoteric oxide because it reacts with acids as well as bases to form corresponding salts. 

SnO2 +4HCl → SnCl4 +2H2O
SnO2 + 2 NaOH → Na2SnO3 + H2O

476 Views

20.

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y-) will be 

  • 275.1 pm

  • 322.5 pm

  • 241.5 pm

  • 241.5 pm


C.

241.5 pm

Radius space ratio space of space NaCl space like space crystal space space equals space straight r to the power of plus over straight r to the power of minus space equals space 0.414
or space straight r to the power of minus space equals space fraction numerator 100 over denominator 0.414 end fraction space equals space 241.5 space pm
621 Views