1.  Identify the wrong statement in the following (AIEEE 2008).

A. CFCs are responsible for ozone layer depletion.

B. Greenhouse effect is responsible for global warming.

C. Ozone layer does not permit I.R. radiation from the sun to reach the earth.

D. Acid rain is mostly because of oxides of ‘N’ and ‘S’.

2.  In the reaction $$2Al_{(s)}+6HCl_{(aq)}\rightarrow 2Al^{3+}_{(aq)}+6Cl^{-}_{(aq)}+3H_{2(g)}$$ (AIEEE 2007)

A. 6L $$HCl_{(aq)}$$ is consumed for every 3L $$H_{2(g)}$$ produced.

B. 33.6L $$H_{2(g)}$$ is produced regardless of temperature and pressure for every mole of Al that reacts.

C. 67.2L $$H_{2(g)}$$ at STP, is produced for every mole Al that reacts.

D. 11.2L $$H_{2(g)}$$ at STP, is produced for every mole $$HCl_{(aq)}$$ consumed.

3.  Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenyl amine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (IIT JEE 2007)

A. 3

B. 4

C. 5

D. 6

Explanation :

The redox reaction between potassiumdichromate and Mohr’s salt is :

$$6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_{2}O$$

4.  Which has maximum number of atoms ? (IIT JEE 2003)

A. 24g of C (12)

B. 56g of Fe(56)

C. 27g of Al (27)

D. 108g of Ag (108)

Explanation :

Number of particles $$\alpha$$ Number of moles

$$\therefore$$ No. of moles of carbon = 24/12 = 2

5.  What volume of hydrogen gas at 273K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) form the reduction of boron trichloride by hydrogen (AIEEE 2003)

A. 89.6L

B. 67.2L

C. 44.8L

D. 22.4L

Explanation :

$$2BCl_{3}+3H_{2}\rightarrow 2B+6HCl$$

$$\therefore V=\frac{nRT}{P}$$

$$=\frac{3\times 0.0821\times 273}{1}$$

$$=67.2L$$

6.  In an organic compound of molar mass 108 g/mol, C,H and N atoms are present in 9:1:3.5 by weight Molecular formula can be (AIEEE 2002)

A. $$C_{6}H_{8}N_{2}$$

B. $$C_{6}H_{10}N$$

C. $$C_{5}H_{6}N_{3}$$

D. $$C_{4}H_{18}N_{3}$$

Explanation :

mass of carbon = $$\frac{9}{13.5}\times 108$$ = 72g

$$\therefore$$ No. of carbon atoms = 72/12 = 6

similarly, no of  H and N atoms are 8 and 2 respectively.

7.  Number of atoms in 560 g of Fe (atomic mass = 56) is (AIEEE 2002)

A. Twice that of 70 g N.

B. Half that of 20 g H

C. Both A and B

D. None of these

8.  In the standardization of $$Na_{2}S_{3}O_{2}$$ using $$K_{2}Cr_{2}O_{7}$$ by iodometry, the equivalent weight of $$K_{2}Cr_{2}O_{7}$$ (IIT JEE 2001)

A. $$\frac{Molar\ mass}{2}$$

B. $$\frac{Molar\ mass}{6}$$

C. $$\frac{Molar\ mass}{3}$$

D. Same as molar mass

Explanation :

During the reaction, $$Cr_{2}O_{7}^{2-}$$ changes to $$Cr^{3+}$$ . Hence the change in oxidation number of Cr is 6.

$$\therefore Equivalent\ weight=\frac{Molar\ mass}{6}$$

9.

Mixture X=0.02 mole of $$\left [ Co(NH_{3})_{5}SO_{4} \right ]Br$$ and 0.02 mole of $$\left [ Co(NH_{3})_{5}Br \right ]SO_{4}$$ was prepared in 2L of Solution

1L of mixture X + excess $$AgNO_{3}\rightarrow Y$$

1L of mixture X + excess $$BaCl_{2}\rightarrow Z$$

Number of mole of Y and Z are            (IIT JEE 2003)

A. 0.01, 0.01

B. 0.02, 0.01

C. 0.01,0.02

D. 0.02, 0.02

Explanation :

In 2L solution, there are 0.02 mol $$Br^{-}$$ ions and 0.02 mole $$SO_{4}^{2-}$$

$$\therefore$$ 1 L of mixture X contains 0.01 mol $$Br^{-}$$ and 0.01 mol $$SO_{4}^{2-}$$ ions.

Hence, Y= 0.01 mol AgBr

Z= 0.01 mol $$BaSO_{4}$$

10.  How many moles of electron weight one kilogram ? (IIT JEE 2002)

A. $$6.023\times 10^{23}$$

B. $$\frac{1}{9.108}\times 10^{31}$$

C. $$\frac{6.023}{9.108}\times 10^{54}$$

D. $$\frac{1}{9.108\times 6.023}\times 10^{8}$$

Mass of an electron $$= 9.108\times 10^{-31}kg$$
No of electron's in 1 Kg $$=\frac{1}{9.108\times 10^{-31}}$$
$$=\frac{1}{9.108\times 10^{-31}\times 6.023\times 10^{23}\ mol^{-1}}$$
$$=\frac{10^{8}}{9.108\times 6.023}\ mol$$