﻿ pH of a solution of 10 mL 1N sodium acetate and 50 mL 2N acetic acid (Ka = 1.8 x 10-5), is approximately : from Chemistry NEET Year 2003 Free Solved Previous Year Papers

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# 1.pH of a solution of 10 mL 1N sodium acetate and 50 mL 2N acetic acid (Ka = 1.8 x 10-5), is approximately :4 5 6 7

A.

4

Given, Ka = 1.8 x 10-5

[salt] = 1 x 10 = 10 millimol

[Acid] = 2 x 50 = 100 millimol

From Henderson's equation,

pH = pKa + log $\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

= - log (1.8 x 10-5) + log $\frac{10}{100}$

= - log 1.8 + 5 + log 10-1

= - 0.2553 + 5 - 1 = 3.7447

= 4

2.

The solution obtained by adding water to alcohol, shows :

• +ve deviation from Raoult's law

• ideal behaviour

• -ve deviation from Raoult's law

• application of Henry's law

A.

+ve deviation from Raoult's law

For the solution of alcohol and water, A ... B attractions, are lesser than B...B and A ... A attractions, hence the vapour pressure of solution is greater than expected. i.e., this solution shows positive deviation from Raoult's law ($∆$Vm= +ve, $∆$Hm= +ve).

3.

How many neutrons are there in nucleus 'X', if it gives 7N14 after two successive $\mathrm{\beta }$ - particle emissions ?

• 7

• 10

• 14

• 09

D.

09

The atomic number of radioactive element increases by one by the emission of 1 $\mathrm{\beta }$ - particle. Hence, after the emission of 2 $\mathrm{\beta }$ -particles, there will be an increase of 2 units in atomic number.

Hence, 5X14  7N14

Hence, the number of neutrons in parent element (X) = 14 - 5 = 9

4.

What will be the normality of a solution, containing 4.9 g. H3PO4 dissolved in 500 ml water ?

• 0.3

• 1.0

• 3.0

• 0.1

A.

0.3

Given that, volume = 500 mL = 0.5 litre

weight of solute (w) = 4.9 g

equivalent weight of

H3PO4

= $\frac{98}{3}$

Hence, normality of solution (N)

$\frac{4.9}{\frac{98}{3}×0.5}$ = 0.3N

5.

2.0 molar solution is obtained, when 0.5 mole solute is dissolved in :

• 250 ml solvent

• 250 g solvent

• 250 ml solution

• 1000 ml solvent

B.

250 g solvent

We know that,

Molality =

$\therefore$ 2.0 =

$\therefore$ weight of solvent = $\frac{0.5}{2.0}$ = 0.250 Kg = 250 g

6.

Which particle can be used to change 13Al27 into 15P30 ?

• Neutron

• $\mathrm{\alpha }$ - particle

• Proton

• Deuteron

B.

$\mathrm{\alpha }$ - particle

$\mathrm{\alpha }$ - particle can be used to convert 13Al27 into 15P30 .

13Al272He4 $\to$ 15P300n1

7.

According to Bohr, the angular momentum of an electron, in any orbit, should be :

• h/2$\mathrm{\pi }$

• nh/2$\mathrm{\pi }$

• h/mv

• 2$\mathrm{\pi }$/nh

B.

nh/2$\mathrm{\pi }$

According to Bohr's atomic model, electrons can revolve only in those orbits, where its angular momentum is a multiple of h/2$\mathrm{\pi }$ ie equal to nh/2$\mathrm{\pi }$ (where n is an integer)

8.

The pH is less than 7, of the solution of :

• FeCl3

• NaCN

• NaOH

• NaCl

A.

FeCl3

FeCl3 is a salt of strong acid and weak base, hence on hydrolysis, it gives acidic solution. Hence, the pH value of solution will be less than 7

9.

The energy equivalent to 1 amu., is :

• 931.1 MeV

• 1.492 x 1013 erg

• 1000 J

• 107 erg

A.

931.1 MeV

From Einstein equation, E = mc2

$\therefore$ Energy equivalent to 1 amu

(E)= 1 a.m.u. x (3 x 108 m/s)2

= 1.66 x 10-27 Kg x 9 x 1016 m2/s2

= 14.94 x 10-11 joule

(Kg m2/s2 = joule)

= 931.1 MeV

10.

The values of Ksp for CuS, Ag2S and HgS are 10-31, 1042, 10-54 respectively. The correct order of their solubility in water is :

• AgS > HgS > CuS

• HgS > Cus > Ag2S

• HgS > Ag2S > Cus

• Ag2S > CuS > HgS

D.

Ag2S > CuS > HgS

Solubility product (Ksp) of CuS = 10-31

$\therefore$ Solubility of Cus = $\sqrt{{\mathrm{K}}_{\mathrm{sp}}}$ = $\sqrt{{10}^{-31}}$

= 3.16 x 10-16 mol/lit

Similarly, solubility of Ag2S

$3\sqrt{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}$

($\because$ for Ag2S, 4s3 = Ksp)

$\sqrt{\frac{{10}^{-42}}{4}}$

= 6.3 x 10-15 mol/litre

Solubility of HgS = $\sqrt{{\mathrm{K}}_{\mathrm{sp}}}$

$\sqrt{{10}^{-54}}$

= 10-27 mol/lit

Hence, the correct order of solubility is

Ag2S > Cus > HgS