NEET Class 12

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 Multiple Choice QuestionsLong Answer Type



In 'x' mL 0.3 N HCl, the addition of 200 mL distilled water or addition of 100 mL 0.1 N NaOH, gives the same final acid strength. Determine 'x'.

Case I

When 200 mL distilled water is added to x mL solution

x X 0.3 = (x + 200) X y

Here, y = final normality or concentration.

 y = 0.3x(x + 200)

Case II

Number of equivalents of HCl = 0.3 × x1000

Number of equivalents of NaOH = 100 × 0.11000 = 0.01

Number of equivalents of HCl remained after addition of NaOH = 0.3 x1000- 0.01

 Concentration = 0.3 x1000- 0.01100 + x x 1000

As final acid strength is same in both cases 0.3 x1000-0.01 1000100 + x = 0.3 x200 + x

or (0.3 x -10) x (200 + x) = (0.3x)(100 + x)

or 60 x -2000 + 0.3x2 - l0x = 30x + 0.3x2

or 20x = 2000 or x = 100 ml


Compound A treated with NaNH2 followed by CH3CH2Br gave compound B. Partial hydrogenation of compound B produced compound C, which on ozonolysis gave a carbonyl compound D, (C2H5O). Compound D did not respond to the iodoform test with I2/KI and NaOH. Find out the structures of A, B, C and D.


The bacterial growth follows the rate law, dN/dt = kN where k is a constant and 'N' is the number of bacteria cell at any time. If the population of bacteria (number of cell) is doubled in 5 min find the time in which the population will be eight times of the initial one?


An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen's reagent and undergoes Cannizaro reaction. On vigorous oxidation, it gives a dicarboxylic acid which is used in the preparation of terylene. Identify the organic compound.


Deep blue CuSO4.5H2O is converted to a bluish-white salt at 100oC. At 250oC and 750oC it is then transformed to white powder and black material respectively. Identify the salts.