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QUESTION: 1

At S.T.P. the density of CCl_{4} vapours in g/L will be nearest to: [1988]

Solution:

1 mol CCl_{4} vapour = 12 + 4 × 35.5 = 154 g ≡ 22.4 L at STP

∴ Density =

QUESTION: 2

One litre hard water contains 12.00 mg Mg^{2+}.Mili-equivalents of washing soda required to remove its hardness is: [1988]

Solution:

QUESTION: 3

1 c.c. N_{2}O at NTP contains :

Solution:

As we know, 22400 cc of N_{2}O contain 6.02 x 10^{23} molecules

Since in N_{2}O molecule there are 3 atoms

No. of electrons in a molecule of N_{2}O = 7 + 1 + 8 = 22

QUESTION: 4

A metal oxide has the formula Z_{2}O_{3}. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is [1989]

Solution:

The reactioin may given as

Z_{2} O_{3} + 3H_{2} —→ 2Z + 3H_{2}O

0.1596 g of Z_{2}O_{3} react with H_{2}

= 6 mg = 0.006 g

∴ 1 g of H_{2} react with

26.6g of Z_{2}O_{3}

∴ Eq. wt. of Z_{2}O_{3 }= 26.6 (from the elefinition of eq. wt.) Eq. wt. of Z + Eq. wt. of O = E + 8 = 26.6

⇒ Eq. wt. of Z = 26.6 – 8 = 18.6 Valency of metal in Z_{2}O_{3} = 3

Eq. wt.of metal

∴ At. wt. of Z = 18.6 × 3 = 55.8

QUESTION: 5

Ratio of C_{p} and C_{v} of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be [1989]

Solution:

C_{p} / C_{v} = 1.4 shows that the gas is diatomic. 22.4 litre at NTP ≡ 6.02 × 10^{23} molecules 11.2 L at NTP = 3.01 × 10^{23 }molecules

= 3.01 × 10^{23} × 2 atoms = 6.02 × 10^{23 }atoms

QUESTION: 6

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Solution:

C_{2}H_{4} + 3 O_{2} —→ 2CO_{2} + 2H_{2}O 28 kg 96 kg

∵ 28 kg of C_{2}H_{4 }undergo complete combustion by = 96 kg of O_{2}

∴ 2.8 kg of C_{2}H_{4} undergo complete combustion by = 9.6 g of O_{2}.

QUESTION: 7

The number of gram molecules of oxygen in 6.02 × 10^{24 }CO molecules is [1990]

Solution:

6.02 × 10^{23} molecules of CO =1mole of CO 6.02 × 10^{24} CO molecules = 10 moles CO = 10 g atoms of O = 5 g molecules of O_{2}

QUESTION: 8

The number of oxygen atoms in 4.4 g of CO_{2} is

Solution:

4.4 g CO_{2} =0.1 mol CO_{2}

(mol. wt. of CO_{2} = 44)

= 6 × 10^{22} molecules

= 2 × 6 × 10^{22} atoms of O

QUESTION: 9

Boron has two stable isotopes, ^{10}B (19%) and ^{11}B (81%). Average atomic weight for boron in the periodic table is [1990]

Solution:

Average atomic mass

Where R.A. = relative abundance

M.No = Mass number

QUESTION: 10

The molecular weight of O_{2} and SO_{2} are 32 and 64 respectively. At 15°C and 150 mm Hg pressure, one litre of O_{2} contains ‘N’ molecules. The number of molecules in two litres of SO_{2} under the same conditions of temperature and pressure will be : [1990]

Solution:

A ccording to Avogadro's law "equal volumes of all gases contain equal numbers of molecules under similar conditions of temperature and pressure". Thus if 1 L of one gas contains N molecules, 2 L of any gas under the same conditions will contain 2N molecules.

QUESTION: 11

A 5 molar solution of H_{2}SO_{4} is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be : [1991]

Solution:

5 MH_{2}SO_{4} = 10 N H_{2}SO_{4}, (∵Basicity of H_{2}SO_{4} = 2) N_{1}V_{1} = N_{2}V_{2},

10 × 1 = N_{2} × 10 or N_{2 }= 1 N

QUESTION: 12

If N_{A} is Avogadro’s number then number of valence electrons in 4.2g of nitride ions (N^{3–}) is

Solution:

No of moles of nitride ion

= 0.3 mol 0.3 x N_{A} nitride ions.

Valence electrons = 8 × 0.3 N_{A} = 2.4 N_{A }(5 + 3 due to charge). One N^{3–} ion contains 8 valence electrons.

QUESTION: 13

In the final answer of the expression

the number of significant figures is : [1994]

Solution:

On calculation we find

As the least precise number contains 3 significant figures therefore answers should also contains 3 significant figures.

QUESTION: 14

The weight of one molecule of a compound C _{60} H_{122} is [1995]

Solution:

Molecular weight of C_{60}H_{122} = (12 × 60) + 122 = 842.

Therefore weight of one molecule

QUESTION: 15

The percentage weight of Zn in white vitriol [ZnSO_{4}.7H_{2}O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]

Solution:

Molecular weight of ZnSO_{4} .7H _{2}O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287.

∴ percentage mass of zinc (Zn)

QUESTION: 16

Liquid benzene (C_{6}H_{6}) burns in oxygen according to the equation 2C_{6}H_{6}(l ) + 15O_{2}(g) —→ 12CO_{2}(g) 6H_{2}O(g) How many litres of O_{2} at STP are needed to complete the combustion of 39 g of liquid benzene?(Mol. wt. of O_{2} = 32, C_{6}H_{6} = 78) [1996]

Solution:

156 gm of benzene required oxygen = 15 × 22.4 litre

∴ 1 gm of benzene required oxygen

∴ 39 gm of Benzene required oxygen

QUESTION: 17

An organic compound containing C, H and N gave the following analysis : C = 40% ; H = 13.33% ; N = 46.67%Its empirical formula would be [1998]

Solution:

As the sum of the percentage of C, H & N is 100. Thus it does not contains O atom.

Hence empirical formula = CH_{4}N

QUESTION: 18

The number of significant figures for the three numbers 161 cm, 0.161 cm, 0.0161 cm are[1998]

Solution:

We know that all non -zero digits are significant and the zeros at the beginning of a number are not significant. Therefore number 161 cm, 0.161 cm and 0.0161cm have 3, 3 and 3 significant figures respectively.

QUESTION: 19

Haemoglobin contains 0.334% of iron by weight.The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe is 56) present in one molecule of haemoglobin are [1998]

Solution:

Given : Percentage of the iron = 0.334%; Molecular weight of the haemoglobin = 67200 and atomic weight of the iron = 56. We know that the number of iron atoms

QUESTION: 20

In the reaction 4 NH_{3} (g) + 5O_{2}(g) → 4NO(g) + 6H_{2}O(l) When 1 mole of ammonia and 1 mole of O_{2} are made to react to completion, [1998]

Solution:

According to Stoichiometry they should react as follow

Thus for 1 mole of O_{2} only 0.8 mole of NH_{3} is consumed. Hence O_{2} is consumed completely.

QUESTION: 21

An organic compound containing C, H and O gave on analysis C – 40% and H – 6.66%. Its empirical formula would be [1999, 94]

Solution:

(% of O in organic compound Table for empirical formula : = 100 – (40 + 6.66 ) = 53.34 % )

Empirical formula of organic compound = CH_{2}O.

QUESTION: 22

Assuming fully decomposed, the volume of CO_{2} released at STP on heating 9.85 g of BaCO_{3} (Atomic mass, Ba = 137) will be [2000]

Solution:

BaCO_{3} → BaO + CO_{2 }

197 gm

197 gm of BaCO_{3 }released carbon dioxide = 22.4 litre at STP

∴ 1 gm of BaCO_{3} released carbon dioxide

∴ 9.85 gm of BaCO_{3} released carbon dioxide

= 1.12 litre

QUESTION: 23

Specific volume of cylindrical virus particle is 6.02 × 10^{–2} cc/gm. whose radius and length 7 Å & 10 Å respectively. If NA = 6.02 × 10^{23}, find molecular weight of virus [2001]

Solution:

Specific volume (volume of 1 gm) of cylindrical virus particle = 6.02 × 10^{–2} cc/gm

Radius of virus (r) = 7 Å = 7 × 10^{–8} cm

Length of virus = 10 × 10^{–8 }cm

Volume of virus

= 154 × 10^{–23} cc

Wt. of one virus particle

∴ Mol. wt. of virus = Wt. of N_{A }particle

= 15400 g/mol = 15.4 kg/mole

QUESTION: 24

Percentage of Se in per oxidase anhydrase enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxidase anhydrase enzyme is [2001]

Solution:

Suppose the mol. wt. of enzyme = x

Given 100g of enzyme wt of Se = 0.5 gm

∴ In xg of enzyme wt. of Se

Hence

∴ x = 15680 = 1.568 × 10^{4}

QUESTION: 25

Which has maximum number of molecules?

Solution:

2g of H_{2} means on e mole of H_{2}, hence contains 6.023 × 10^{23 }molecules. Others have less than one mole, so have less no. of molecules.

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