﻿ Equation of Boyle s law is : from Chemistry NEET Year 2006 Free Solved Previous Year Papers

## 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 2006

#### Multiple Choice Questions

1.

For an electron , if the uncertainty in velocity is$∆$v ,the uncertainty in its position ($∆$x) is given by :

• $\frac{\mathrm{hm}}{4\mathrm{\pi }∆\mathrm{\nu }}$

• $\frac{4\mathrm{\pi }}{\mathrm{hm}∆\mathrm{\nu }}$

• $\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

• $\frac{4\mathrm{\pi m}}{\mathrm{h}∆\mathrm{\nu }}$

C.

$\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

According to Heisenberg's uncertainty principle

$∆$x.$∆$p$⩾$$\frac{\mathrm{h}}{4\mathrm{\pi }}$ ($∆$p = m$∆\mathrm{\nu }$)

$∆$x. m$∆$v$⩾$$\frac{\mathrm{h}}{4\mathrm{\pi }}$

$∆$x =$\frac{\mathrm{h}}{4\mathrm{\pi m}∆\mathrm{v}}$

2.

The enthalpy change ($∆$H) for the neutralisationof M HCl by caustic potash in dilute solution at298 K is :

• 68 kJ

• 65 kJ

• 57.3 kJ

• 50 kJ

C.

57.3 kJ

$\underset{\left(\mathrm{strong}\mathrm{acid}\right)}{\mathrm{HCl}}$+$\underset{\left(\mathrm{strong}\mathrm{base}\right)}{\mathrm{KOH}}$$\stackrel{\mathrm{neutralisation}}{⇌}$KCl + H2O

In this reaction , HCl is the strong acid and KOHis the base and it has been found that the heat ofneutralisation of a strong acid with a strong baseis always constant i.e. , 57.3 kJ .

3.

Orbital is :

• circular path around the nucleus in which the electrons revolves

• space around the nucleus where the probability of finding the electron is maximum

• amplitude of electron wave

• none of the above

B.

space around the nucleus where the probability of finding the electron is maximum

An atomic orbital is a three dimensional regionof definite shape around the nucleus where theprobability of finding the electron is maximum .Therefore , an atom has a both characteristicenergy and a characteristic shape .

4.

A radioactive sample is emitting 64 timesradiations than non-hazardous limit .If itshalf-life is 2 hours , after what time it becomesnon- hazardous ?

• 16 h

• 10 h

• 12 h

• 8 h

C.

12 h

We know ,

Nt = Nox${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

where Nt = amount left after expiry of 'n' half lives

No = initial amount

n = number of half lives elapsed

$\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_{\mathrm{o}}}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

$\frac{1}{64}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

${\left(\frac{1}{2}\right)}^{6}$=${\left(\frac{1}{2}\right)}^{\mathrm{n}}$

n = 6

Time taken (T) = t1/2 x n = 2 x 6 = 12 h

5.

A metal surface is exposed to solar radiations :

• the emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations

• the emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiations

• the emitted electrons have zero energy

• the emitted electrons have energy equal to energy of photons of incident light

A.

the emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiations

A metal surface is exposed to solar radiations ,the emitted electrons have energy less thanmaximum value of energy depending uponfreauency of incident radiations .

6.

Ionic compounds are formed most easily with :

• low electron affinity , high ionisation energy

• high electron affinity , low ionisation energy

• low electron affinity , low ionisation energy

• high electron affinity , high ionisation energy

B.

high electron affinity , low ionisation energy

An ionic bond is formed between an electropositive metal and a electronegative non metal .The electropositive metal converts into a cation ,while the electronegative non metal convertsinto a anion .

Thus the formation ofionic bond is favoured by

(I)Low ionisation potential of metal

(II)Greater valtie of electron affinity of nonmetal

(III)Higher value of lattice energy of theresulting ionic compound .

7.

The Ksp of Mg(OH)2 is 1 x 10-12 , 0.01 M Mg(OH)2will precipitate at the limited pH :

• 3

• 9

• 5

• 8

B.

9

Mg(OH)2$⇌$Mg2+ + 2OH-

the solubility product Kspof

Mg(OH)2 = [Mg2+][OH-]2

1 x 10-12 = 0.01 [OH-]2

[OH-]2 = 1 x 10-10

[OH-] = 10-5

We know

[H+][OH-] = 10­-14

[H+][10-5]= 10-14

[H] = 10-14/10-5 = 10-9

pH = - log[H+] = - log 10-9 = 9

8.

Which of the following sequence is correct as perAufbau principle ?

• 3s < 3d < 4s < 4p

• ls < 2p < 4s < 3d

• 2s < 5s < 4p < 5d

• 2s < 5s < 4p < 5f

B.

ls < 2p < 4s < 3d

According to aufbau principle , "sub-shells arefilled with electrons in the increasing order oftheir energies , "i.e. , sub-shell of lower energywill be filled first with electrons .

Thus correct order is -

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p< 5s < 4d < 5p < 6s < 4f < 5d < 6p< 7s < 5f < 6d ... etc

9.

Which of the following transitions haveminimum wavelengths ?

• n4 $\to$ n1

• n2 $\to$ n1

• n4 $\to$ n2

• n3 $\to$ n1

A.

n4 $\to$ n1

E =$\frac{\mathrm{hc}}{\mathrm{\lambda }}$or E$\propto$$\frac{1}{\mathrm{\lambda }}$

Thus a decrease in wave length represents anincrease in energy Forn4$\to$n1transition .These are greater the. energy difference andlesser will be wavelength .Thus , forn4$\to$n1 transition has minimum wavelength .

# 10.Equation of Boyle's law is :$\frac{\mathrm{dP}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{V}}$ $\frac{\mathrm{dP}}{\mathrm{P}}$ = +$\frac{\mathrm{dV}}{\mathrm{V}}$ $\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{dT}}$ $\frac{{\mathrm{d}}^{2}\mathrm{P}}{\mathrm{P}}$ = +$\frac{{\mathrm{d}}^{2}\mathrm{V}}{\mathrm{dT}}$

A.

$\frac{\mathrm{dP}}{\mathrm{P}}$ = -$\frac{\mathrm{dV}}{\mathrm{V}}$

According to Boyle's law , "for a given mass of agas , at constant temperature the volume of a gasis inversely proportional to its pressure" .

V$\propto$$\frac{1}{\mathrm{P}}$

or PV = constant

on differentiating the equation

d(PV) = d (constant)

= PdV + VdP =0

= VdP = - PdV

=$\frac{\mathrm{dP}}{\mathrm{P}}$= -$\frac{\mathrm{dV}}{\mathrm{V}}$