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# NEET Chemistry Solved Question Paper 2013

#### Multiple Choice Questions

1.

If the equilibrium constant of the reaction of weak acid HA with strong base is 109 then pH of 0.l M NaA is :

• 5

• 9

• 7

• 8

B.

9

HA + OH- $⇌$ H2O + A-

K = $\frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]\left[{\mathrm{OH}}^{-}\right]}$

HA $⇌$ H+ + A-

Ka$\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

$\frac{{\mathrm{k}}_{\mathrm{a}}}{\mathrm{K}}$ = [H+][OH-] = Kw

or    Ka = K2K = 10-14 $×$ 109 = 10-5

pKa = 5

A- solution is alkaline due to hydrolysis

$\therefore$ pH = 7 + $\frac{{\mathrm{pK}}_{\mathrm{a}}}{2}$ +

= 7 + $\frac{5}{2}$ +  = 9

2.

Mass of one atom is 6.66 x 10-23 g. Its percentage in an hydride is 95.24. Thus , hydride is :

• MH

• MH2

• MH3

• MH4

B.

MH2

Mass of one atom= 6.66 x 10-23 g

Mass of N0 atoms = 6.66 x 10-23 x 6.02 x 10-23 g = 40 g

Thus , atomic weight of the element = 40

 Element % % at.wt. Ratio M 95.24 2.381 1 H 4.76 4.76 2

Thus , hydride is MH.

3.

1 mol of O2 and x mol of Ne in a 10 L flask at constant temperature exert a pressure of 10 atm. If partial pressure of O2 is 2 atm , moles of Ne in the mixture is :

• 1

• 2

• 4

• 3

C.

4

D.

3

pO2 = ptotal ${\mathrm{p}}_{{\mathrm{o}}_{2}}$

${\mathrm{p}}_{{\mathrm{o}}_{2}}$ =

${\mathrm{p}}_{{\mathrm{o}}_{2}}$ = 2 = 10 x  or 10 = 2 + 2x

x = 8/2 = 4 mol

4.

A 1 L flask contains 32 g O2 gas at 27C. What mass of Omust be released to reduce the pressure in the flask to 12.315 atm?

• 8 g

• 16 g

• 24 g

• 32 g

B.

16 g

pV = nRT = $\frac{\mathrm{w}}{\mathrm{m}}$RT

w = $\frac{\mathrm{pVm}}{\mathrm{RT}}$ =  = 16 g

O2 to be released= 32 - 16 = 16 g

5.

${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (C2H4) , ${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (C2H6) are x1 and x2 kcalmol-1 respectively. Then heat of hydrogenation of C2H4 is :

• x1 + x2

• x1 - x2

• -(x1 - x2)

• x1 + 2x2

C.

-(x1 - x2)

C2H4 + H2 $\to$C2H6

$\mathrm{\Delta }$H = ${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (C2H6) - ${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (C2H4) - ${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (H2)

= x2 - x1 - 0

= x2 - x1

= -(x1 - x2)

6.

NO2 (brown colour gas) exists in equilibrium with N2O4 (colourless gas) as given by chemical equation.

Mixture is slightly brown due to existence of NO2. If pressure is increased :

• colour intensity is increased

• colour intensity is decreased

• colour intensity first increases and then decreases

• No change in colour intensity

D.

No change in colour intensity

By Le-Chateliers principle, due to increase in pressure, equilibrium shifts in the direction of low pressure (in which number of mole decreases) i.e. , forward side.

# 7.For the process ,X (g) + e- $\to$ X- (g) ; $\mathrm{\Delta }$H = xand X- (g) $\to$ X (g) + e- ; $\mathrm{\Delta }$H = yselect correct alternateIonisation energy of x- (g) is y Electron affinity of X (g) is x Electron affinity of X (g) is - y All of the above

D.

All of the above

X (g) + e- $\to$ X- (g) ; $∆$H = x = EA of X (g)

X- (g) $\to$ X (g) + e-$∆$H = y = IE of X- (g)

= - EA of X (g)

Therefore , all options are correct .

8.

0.1 M solution of CH3COOH should be diluted to how many times so that pH is doubled ?

• 4.0 times

• 5.55 x 104 times

• 5.55 x 106 times

• 10-2 times

B.

5.55 x 104 times

pH = $\frac{1}{2}$[pKa - log1] = $\frac{{\mathrm{pK}}_{\mathrm{a}}}{2}$

pH' (twice of pH) = pKa

pKa$\frac{1}{2}$[pKa - logC]

logC = pKa = -logKa

C = Ka = 1.8 x 10-5 M

$\therefore$ Dilution = $\frac{1}{\mathrm{C}}$ =  = 5.55 x 104 times.

9.

Spontaneous adsorption of gas on solid surface is an exothermic process because :

• $\mathrm{\Delta }$H increases for system

• $\mathrm{\Delta }$S increases for gas

• $\mathrm{\Delta }$S decreases for gas

• $\mathrm{\Delta }$G increases for gas

C.

$\mathrm{\Delta }$S decreases for gas

When a gas is adsorbed spontaneously on a solid surface, entropy decreases for gas.

10.

The heat of formation of C12H22O11 (s) , CO2 (g) and (H2O) are - 530 , -94.3 and- 68.3 kcalmol-1 respectively. The amount of C12H22O11 to supply 2700 kcal of energy is :

• 382.70 g

• 832.74 g

• 463.9 g

• 684.0 g

D.

684.0 g

C12H22O11 (s) + 12O2 (g) $\to$ 12CO2 (g) + 11H2O (l)

${\mathrm{\Delta H}}_{\mathrm{comb}}^{°}$ = [12${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$(CO2) + 11${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (H2O)] - ${\mathrm{\Delta H}}_{\mathrm{f}}^{°}$ (C12H22O11)]

= - 1352.9 kcal

Thus, number of moles of C12H22O11 for getting 2700 kcal of heat.

$\frac{2700}{1352.9}$ = 2 mol

= 2 X 242 = 684.0 g