0.126 g of an acid is needed to completely neutralise 20 mL 0.1 (

Subject

Chemistry

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

For same mass of two different ideal gases of molecular weights M1 and M2, Plots of log V vs log p at a given constant temperature are shown. Identify the correct option.

  • M1 > M2

  • M1 = M2

  • M1 < M2

  • Can be predicted only if temeperature is known


2.

Which of the following has the dimension if [ML0T-2]?

  • Coefficient of viscosity

  • Surface tension

  • Vapour pressure

  • Kinetic energy


3.

If the given four electronic configurations,

(i) n = 4, l = 1

(ii) n = 4, l = 0

(iii) n = 3, l = 2

(iv) n = 3, l = 1

are arranged in order of increasing energy, then the order will be

  • (iv) < (ii) < (iii) < (i)

  • (ii) < (iv) < (i) < (iii)

  • (i) < (iii) < (ii) < (iv)

  • (iii) < (i) < (iv) < (ii)


4.

Which of the following sets of quantum numbers represents the 19th electron of Cr(Z = 24)?

  • (4, 1, -1, +12)

  • (4, 0, 0, +12)

  • (3, 2, 0, -12)

  • (3, 2, -2, +12)


Advertisement
Advertisement

5.

0.126 g of an acid is needed to completely neutralise 20 mL 0.1 (N) NaOH solution. The equivalent weight of the acid is

  • 53

  • 40

  • 45

  • 63


D.

63

Since, number of milli equivalents of NaOH (n') = NV = 0.1 × 20 = 2

 Number of equivalent = 0.002

For Neutralisation

Number of equivalent of base (NaOH) = Number of equivalent of acid

0.002 = WE = 0.126E

 E = 0.1260.002 = 63


Advertisement
6.

In a flask, the weight ratio of CH4 (g) and SO2 (g) at 298K and 1 bar is 1 : 2. The ratio of the number of molecules of SO2 (g) and CH4 (g) is

  • 1 : 4

  • 4 : 1

  • 1 : 2

  • 2 : 1


7.

C6H5F18 is a F18  radio-isotope labelled organic compound. F18 decays by positron emission. The product resulting on decay is 

  • C6H5O18

  • C6H5Ar19

  • B12C5H5F

  • C6H5O16


8.

Dissolving NaCN in de-ionised water will result in a solution having

  • pH < 7

  • pH = 7

  • pOH = 7

  • pH > 7


Advertisement
9.

Among Me3N, C5H5N and MeCN (Me = methyl group) the electronegativity of N is in the order

  • MeCN > C5H5N > Me3N

  • C5H5N > Me3N > MeCN

  • Me3N > MeCN > C5H5N

  • Electronegativity same in all


10.

The shape of XeF5- will be

  • square pyramid

  • trigonal bipyramidal

  • planar

  • pentagonal bipyramid


Advertisement