﻿ NEET Chemistry Solved Question Paper 2015 | Previous Year Papers | Zigya

## 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

1.

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

• 319.4 m and X-rays

• 219.3 m and microwave

D.

In the above question,

Frequencyv1368 $×$103 Hz

$\mathrm{\lambda }$ = $\frac{c}{v}$ =  = 219.3 m

(where, $\mathrm{\lambda }$ = wavelength; c = velocity of light; v = frequency)

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

2.

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

D.

200 K

As pressure and volume remains constant

n1T1 = n2T2

n1$\frac{0.4}{4}$ = 0.1; T1 = TK

n2 = ; T2 = (T + 50)K

On putting the values,

$⇒$0.1 x T = 0.08 x (T + 50)

$⇒$0.1T = 0.08 T = 4

$⇒$0.1 T - 0.08 T = 4

$⇒$0.02 T = 4

$⇒$T = 200K

3.

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

C.

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 =

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

Therefore, Increasing order of weight

Fe < Zn < Ag < H2CO3 ·2H2O

4.

What would be the heat released when an aqueous solution containing 0.5 mole of HNOis 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

B.

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.

5.

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+

A.

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+

6.

In an amino acid, the carboxyl group ionises at pKa = 2.34 and ammonium ion at pKa2 = 9.60. The isoelectric point of the amino acid is at pH

• 4.32

• 3.34

• 9.46

• 5.97

D.

5.97

Isoelectric point, pH =

=

= 5.97

7.

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 em = 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

C.

1.92 x 10-3 m

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

Accuracy of velocity = 0.005%

$\therefore$ $∆$v =  = 0.03

According to Heisenberg's uncertainty principle

1.92 $×$ 10-3 m

8.

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

C.

> 1

Polydispersity Index or PDI =

PDI =

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

9.

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.

• 8.57 L

• 2.14 L

• 1.28 L

• 7.51 L

B.

2.14 L

Moles needed = 3.00 L $×$ 1.00 M = 3 mol.

Let's assume, x L of 2.50 M NaOH is added. Then (3 - x) L of 0.40 M NaOH is added (Given). ..... (i)

The number of moles of solute from the more concentrated solution is 2.50, that becomes less concentrated solution is (0.40) (3.00 - x).

Total number of moles = 3.00

$\therefore$ 2.50x + 0.40 (3.00 - x) = 3.00

$⇒$  2.50x + 1.2 - 0.40x = 3.00

$⇒$  2.10x + 1.2 = 3.00

$⇒$  2.10x = 1.8

$⇒$   = 0.857 L of 2.50 M NaOH  ... (ii)

Putting value of x from (ii) in (i), is

3 - 0.857 = 2.14 L of 0.40 M NaOH

Therefore, the amount of B stock solution to be added is 2.14 L.

10.

What percentage of $\mathrm{\beta }$-D-(+-glucopyranose is found at equilibrium in the aqueous solution?

• 64%

• 36%

• $\approx$100%

• $\approx$50%

A.

64%

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