Need NEET 2015 Chemistry Solved Previous Year Question Paper

Find solved questions and answers for the paper 2015 with download option

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 2015

Multiple Choice Questions

What would be the heat released when an aqueous solution containing 0.5 mole of HNO₃is mixed with 0.3 mole of OH^-? (enthalpy of neutralisation is- 57.1 kJ)

28.5 kJ
17.1 kJ
45.7 kJ
1.7 kJ

17.1 kJ

0.5 mole of HNO₃ equivalents to 0.5 mole of H⁺. But only 0.3 mole of OH^- ions are available.

Thus, 0.3 mole of H⁺ ions will combine with 0.3 mole of OH^- ions to form 0.3 mole of H₂O and 0.2 mole of H⁺ ions will remain unreacted.

Heat evolved when 1 mole of H⁺ ions combine with 1 mole of OH^- ions = 57.1 kJ.

Therefore, heat evolved when 0.3 mole of H⁺ ions combine with 0.3 mole of OH^- ions = 0.3x 57.1 = 17.13 kJ.

The Vividh Bharti Station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (Kilohertz). Calculate the wavelength ( $λ$ ) of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

319.4 m and X-rays
319.4 m and radiowave
219.3 m and microwave
219.3 m and radiowave

219.3 m and radiowave

In the above question,

Frequency = v = 1368 $\times$ 10³ Hz

$λ$ = $\frac{c}{v}$ = $\frac{3 \times 10^{8} m / s}{1368 \times 10^{3} s^{- 1}}$ = 219.3 m

(where, $λ$ = wavelength; c = velocity of light; v = frequency)

The range of wavelength of radiowaves is 100 m to radiation is 10000 m.

The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species, K⁺, Ca²⁺, Mg²⁺, Be²⁺?

K⁺< Ca²⁺ < Mg²⁺ < Be²⁺
Be²⁺ < K²⁺ < Ca²⁺ < Mg²⁺
Mg²⁺< Be²⁺ < K⁺ < Ca²⁺
Ca²⁺ < Mg²⁺ < Be⁺ < K⁺

K⁺< Ca²⁺ < Mg²⁺ < Be²⁺

High charge and small size of the cations increases polarisation. As the size of the given cations decreases as

K⁺> Ca²⁺ > Mg²⁺ > Be²⁺

Thus, polarising power increases as

K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

Which of the following is correctly arranged in order of increasing weight?

0.0105 equivalent of H₂CO₂ ·2H₂O < 0.625 g of Fe < 0.006 g atom of Ag <6.0 x 10²¹ atoms of Zn
0.625 g of Fe < 0.0105 equivalent of H₂C₂O₄ ·2H₂O < 6.0 x 10²¹ atoms of Zn < 0.006 g atom of Ag
0.625 g of Fe < 6.0 x 10²¹ atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H₂C₂O₄ ·2H₂O
0.0105 equivalent of H₂C₂O₄. 2H2O < 0.006 g atom of Ag <6.0 x 10²¹ atoms of Zn < 0.625 g of Fe

0.625 g of Fe < 6.0 x 10²¹ atoms of Zn < 0.006 g atom of Ag < 0.0105 equivalent of H₂C₂O₄ ·2H₂O

0.625 g Fe

0.0105 eq of H₂C₂O₄. 2H₂O = 0.0105 x 63

= 0.66 g (Eq. wt. = 63)

6 x 10²¹ atoms of Zn = $\frac{6 \times 10^{21}}{6.023 \times 10^{23}} \times 65 = 0.647 g$

0.006 g atom of Ag= 0.006 x 108= 0.648 g

Therefore, Increasing order of weight

Fe < Zn < Ag < H₂CO₃ ·2H₂O

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is

(Given, h= 6.6 x 10^-34 kg m² s^-1, mass of electron e_m = 9.1 x 10^-31 kg)

215 x 10^-3 m
2.78 x 10^-3 m
1.92 x 10^-3 m
3.24 x 10^-3 m

1.92 x 10^-3 m

Given that velocity of electron, v = 600 m/s

Accuracy of velocity = 0.005%

$∴$ $∆$ v = $\frac{600 \times 0.005}{100}$ = 0.03

According to Heisenberg's uncertainty principle

$∆ x . m ∆ v \geq \frac{h}{4 π}$

$∆ x = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 9.1 \times 10^{- 31} \times 0.03}$

$∆ x =$ 1.92 $\times$ 10^-3 m

What percentage of $β$ -D-(+-glucopyranose is found at equilibrium in the aqueous solution?

64%
36%
$\approx$ 100%
$\approx$ 50%

64%

Ordinary glucose is $α$ -glucose, with a fresh aqueous solution has specific rotation, [ $α$ ]_D = + 110°. On keeping the solution for sometimes, $α$ -glucose slowly changes into an equilibrium mixture of $α$ -glucose (36%) and $β$ -glucose (64%) and the mixture has specific rotation+ 52.5°.

The number average molar mass and mass average molar mass of a polymer are respectively 30,000 and 40,000. The
polydispersity index (PDI) of the polymer is

-1
0
> 1
< 1

> 1

Polydispersity Index or PDI = $\frac{Mass average molar mass}{Number average molar mass} = \frac{M_{w}}{M_{n}}$

PDI = $\frac{40000}{30000} = 1.3 > 1$

Therefore, polydispersity index or pDI of the polymer is > 1.

0.40 g of helium in a bulb at a temperature of T K had a pressure of p atm. When the bulb was immersed in water bath at temperature 50 K more than the first one, 0.08 g of gas had to be removed to restore the original pressure. T is

500 K
400 K
600 K
200 K

200 K

As pressure and volume remains constant

n₁T₁ = n₂T₂

n₁ = $\frac{0.4}{4}$ = 0.1; T₁ = TK

n2 = $\frac{0.40 - 0.08}{4} = 0.08$ ; T₂ = (T + 50)K

On putting the values,

$\Rightarrow$ 0.1 x T = 0.08 x (T + 50)

$\Rightarrow$ 0.1T = 0.08 T = 4

$\Rightarrow$ 0.1 T - 0.08 T = 4

$\Rightarrow$ 0.02 T = 4

$\Rightarrow$ T = 200K

In an amino acid, the carboxyl group ionises at pK_a = 2.34 and ammonium ion at pK_a₂ = 9.60. The isoelectric point of the amino acid is at pH

4.32
3.34
9.46
5.97

5.97

Isoelectric point, pH = $\frac{{pK}_{a_{1}} + {pK}_{a_{2}}}{2}$

= $\frac{234 + 9.60}{2}$

= 5.97

10.

For preparing 3.00 L of 1M NaOH by mixing portions of two stock solutions (A and B) of 2.50 M NaOH and 0.40 M NaOH respectively. Find out the amount of B stock solution (in L) added.