## 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.

# NEET Chemistry Solved Question Paper 2011

41.

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.

The Deep blue CuSO4.5H2O is blue vitriol. In its structure, one molecule of water is hydrogen-bonded from 4 sides and thus is released with more difficulty than the rest four water molecules.

42.

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.

The positive Brady's test (i.e., reaction with 2, 4-dinitrophenylhydrazine) indicates that the organic compound is carbonyl compound and Tollen's and Cannizaro reactions indicate that it is an aldehyde without $\mathrm{\alpha }$-hydrogen atom. As the end product is terephthalic acid, the compound must be p-ethyl benzaldehyde.

43.

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.

$\underset{\left(\mathrm{A}\right)}{{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{C}\equiv \mathrm{CH}}$ $\underset{-{\mathrm{NH}}_{3}}{\overset{{\mathrm{NaNH}}_{2}}{\to }}$ CH3CH2C$\equiv \stackrel{-}{\mathrm{C}}\stackrel{+}{\mathrm{Na}}$ $\underset{\mathrm{hydrogenation}}{\overset{\mathrm{partial}}{\to }}$ $\underset{\left(\mathrm{C}\right)}{{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{CH}\equiv {\mathrm{CHCH}}_{2}{\mathrm{CH}}_{3}}$  $\underset{\left(\mathrm{D}\right)}{2{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{CHO}}$

As compound (D) does not have CH3-C=O group, it does not give iodoform reaction.

# 44.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.

$\therefore$ y =

Case II

Number of equivalents of HCl =

Number of equivalents of NaOH =  = 0.01

Number of equivalents of HCl remained after addition of NaOH =

$\therefore$ Concentration =  x 1000

As final acid strength is same in both cases  =

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

45.

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?

As dN/dt = kN (It is a Ist order reaction)

Integrated rate equation for Ist order reaction

k = 2.303/t log N/No

$\because$when t = 5 min, N = 2No

$\therefore$k = 2.303/5 log 2No/No

or k = 2.303/5 log 2

for N = 8No , t = ?

$\therefore$t = 2.303/k log 8No/No

By putting the value of k

t =  log 8No/N

or t =  = 15 min