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Some Basic Concepts of Chemistry

NEET Chemistry · 99 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
A CHO
B CH 4 O
C CH 3 O
D CH 2 O
Correct Answer
Option C
Solution

<table class=tg> <tbody><tr> <th class=tg-s6z2>Element</th> <th class=tg-s6z2>%</th> <th class=tg-s6z2>Atomic mass</th> <th class=tg-s6z2>mole ratio</th> <th class=tg-s6z2>simple ratio</th> </tr> <tr> <td class=tg-s6z2>C</td> <td class=tg-s6z2>38.71</td> <td class=tg-s6z2>12</td> <td class=tg-s6z2>

38.7112=3.22{{38.71} \over {12}} = 3.22

</td> <td class=tg-s6z2>

3.223.22=1{{3.22} \over {3.22}} = 1

</td> </tr> <tr> <td class=tg-s6z2>H</td> <td class=tg-s6z2>9.67</td> <td class=tg-s6z2>1</td> <td class=tg-s6z2>

9.671=9.67{{9.67} \over 1} = 9.67

</td> <td class=tg-s6z2>

9.673.22=1{{9.67} \over {3.22}} = 1

</td> </tr> <tr> <td class=tg-s6z2>O</td> <td class=tg-s6z2>51.62</td> <td class=tg-s6z2>16</td> <td class=tg-s6z2>

51.6216=3.22{{51.62} \over {16}} = 3.22

</td> <td class=tg-s6z2>

3.223.22=1{{3.22} \over {3.22}} = 1

</td> </tr> </tbody></table><br> Hence empirical formula of the compound would be CH<sub>3</sub>O

Q42
What volume of oxygen gas (O 2 ) measuread at 0 o C and 1 atm, is needed to burn completely 1 L of propane gas (C 3 H 8 ) measured under the same conditions ?
A 5 L
B 10 L
C 7 L
D 6 L
Correct Answer
Option A
Solution

<table class=tg> <tbody><tr> <th class=tg-s6z2>C<sub>3</sub>H<sub>8</sub></th> <th class=tg-s6z2>+</th> <th class=tg-s6z2>5O<sub>2</sub></th> <th class=tg-s6z2>→\to</th> <th class=tg-s6z2>3CO<sub>2</sub></th> <th class=tg-baqh>+</th> <th class=tg-baqh>4H<sub>2</sub>O</th> <th class=tg-baqh>(balanced equation)</th> </tr> <tr> <td class=tg-s6z2>1 vol.</td> <td class=tg-s6z2></td> <td class=tg-s6z2>5 vol.</td> <td class=tg-s6z2></td> <td class=tg-s6z2>3 vol.</td> <td class=tg-baqh></td> <td class=tg-baqh>4 vol.</td> <td class=tg-baqh></td> </tr> </tbody></table><br> According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O<sub>2</sub> to burn completely.

Q43
Concentrated aqueous sulphuric acid is 98% H 2 SO 4 by mass and has a density of 1.80 g mL −-1 . Volume of acid required to make one litre of 0.1 M H 2 SO 4 solution is
A 16.65 mL
B 22.20 mL
C 5.55 mL
D 11.10 mL
Correct Answer
Option C
Solution

Normality =

98×1.8×1049{{98 \times 1.8 \times 10} \over {49}}

= 36 N N 2 = 0.1 ×\times 2 = 0.2 N N 2 V 2 = N 1 V 1 ⇒\Rightarrow 36 ×\times V = 0.2 ×\times 1000 ⇒\Rightarrow V =

0.2×100036{{0.2 \times 1000} \over {36}}

= 5.55 mL

Q44
An element, X has the following isotopic composition: 200 X : 90% 199 X : 8.0% 202 X : 2.0% The weighted average atomic mass of the naturally occuring element X is closest to
A 201 amu
B 202 amu
C 199 amu
D 200 amu
Correct Answer
Option D
Solution

Average isotope mass of X =

200×90+199×8+202×290+8+2{{200 \times 90 + 199 \times 8 + 202 \times 2} \over {90 + 8 + 2}}

=

18000+1592+404100{{18000 + 1592 + 404} \over {100}}

= 199.96 a.m.u ≈\approx 200 a.m.u

Q45
The mole fraction of the solute in one molal aqueous solution is
A 0.009
B 0.018
C 0.027
D 0.036
Correct Answer
Option B
Solution

One molal solution means one mole solute present in 1 kg (1000 g) solvent. ∴\therefore mole of solute = 1 Mole of solvent (H 2 O) =

100018{{1000} \over {18}}

X solute =

11+100018{1 \over {1 + {{1000} \over {18}}}}

= 0.018

Q46
The maximum number of molecules is present in
A 15 L of H 2 gas at STP
B 5 L of N 2 gas at STP
C 0.5 g of H 2 gas
D 10 g of O 2 gas
Correct Answer
Option A
Solution

At STP, 22.4 L H 2 = 6.023 ×\times 10 23 molecules 15 L H 2 =

6.023×1023×1522.4{{6.023 \times {{10}^{23}} \times 15} \over {22.4}}

=

4.033×10234.033 \times {10^{23}}

5 L N 2 =

6.023×1023×522.4{{6.023 \times {{10}^{23}} \times 5} \over {22.4}}

= 1.344 ×\times 10 23 2 g of H 2 = 6.023 ×\times 10 23 0.5 g H 2 =

6.023×1023×0.52{{6.023 \times {{10}^{23}} \times 0.5} \over {2}}

= 1.505 ×\times 10 23 32 g O 2 = 6.023 ×\times 10 23 10 g O 2 =

6.023×1023×1032{{6.023 \times {{10}^{23}} \times 10} \over {32}}

= 1.882 ×\times 10 23

Q47
2.5 litre of 1 M NaOH solution is mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity of resultant solution.
A 0.80 M
B 1.0 M
C 0.73 M
D 0.50 M
Correct Answer
Option C
Solution

From molarity equation M 1 V 1 + M 2 V 2 = MV ⇒\Rightarrow 1× 2.5 + 0.5 × 3 = M × 5.5 ⇒\Rightarrow M =

45.5{4 \over {5.5}}

= 0.73M

Q48
The percentage of C, H and N in an organic compound are 40%, 13.3% and 46.7% respectively then empirical formula is
A C 3 H 13 N 3
B CH 2 N
C CH 4 N
D CH 6 N
Correct Answer
Option C
Solution

<table class=tg> <tbody><tr> <th class=tg-nrix></th> <th class=tg-nrix>Element</th> <th class=tg-nrix>%</th> <th class=tg-nrix>Atomic<br>mass</th> <th class=tg-nrix>Relative<br>no. of atoms</th> <th class=tg-baqh>Simplest<br>ratio of atoms</th> </tr> <tr> <td class=tg-nrix>1</td> <td class=tg-nrix>C</td> <td class=tg-nrix>40</td> <td class=tg-nrix>12</td> <td class=tg-nrix>

4012{{40} \over {12}}

= 3.33</td> <td class=tg-baqh>

3.333.3{{3.33} \over {3.3}}

= 1</td> </tr> <tr> <td class=tg-baqh>2</td> <td class=tg-baqh>H</td> <td class=tg-baqh>13.3</td> <td class=tg-baqh>1</td> <td class=tg-baqh>

13.31{{13.3} \over 1}

= 13.3</td> <td class=tg-baqh>

13.33.3{{13.3} \over {3.3}}

= 4</td> </tr> <tr> <td class=tg-baqh>2</td> <td class=tg-baqh>N</td> <td class=tg-baqh>46.7</td> <td class=tg-baqh>14</td> <td class=tg-baqh>

46.714{{46.7} \over {14}}

= 3.3</td> <td class=tg-baqh>

3.33.3{{3.3} \over {3.3}}

= 1</td> </tr> </tbody></table> <br><br>∴\therefore The empirical formula is CH<sub>4</sub>N.

Q49
Which has maximum molecules?
A 7 g N 2
B 2 g H 2
C 16 g NO 2
D 16 g O 2
Correct Answer
Option B
Solution

1 mole = 6.023 ×\times 10 23 number of molecules. 1 g mole of O 2 = 32 g of O 2 ⇒\Rightarrow 16 g of O 2 = 0.5 g mole of O 2 1 g mole of N 2 = 28 g of N 2 ⇒\Rightarrow 7 g of N 2 = 0.25 g mole of N 2 1 g mole of H 2 = 2 g of H 2 ⇒\Rightarrow 2 g of H 2 = 1 g mole of H 2 1 g of NO 2 = 14 + 16 ×\times 2 = 46 ⇒\Rightarrow 16 g of NO 2 = 0.35 mole NO 2

Q50
Specific volume of cylinfrical virus particle is 6.02 × \times 10 −-2 cc/g whose radius and length are 7 A∘\mathop A\limits^ \circ and 10 A∘\mathop A\limits^ \circ respectively. If NA = 6.02 × \times 10 23 , find molecular weight of virus.
A 15.4 kg/mol
B 1.54 × \times 10 4 kg/mol
C 3.08 × \times 10 4 kg/mol
D 3.08 × \times 10 3 kg/mol
Correct Answer
Option A
Solution

Specific volume ( vol. of 1 g) cylindrical virus particle = 6.02 ×\times 10 -2 cc/g Radius of virus, r = 7 Å = 7 ×\times 10 -8 cm Volume of virus =

πr2l\pi {r^2}l

=

227×(7×10−8)×10×10−8{{22} \over 7} \times (7 \times {10^{ - 8}}) \times 10 \times {10^{ - 8}}

= 154 ×\times 10 -23 cc wt. of one virus particle =

VolumeSpecific  Volume{{Volume} \over {Specific\,\,Volume}}

⇒\Rightarrow

1.54×10−236.02×10−2g{{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}}g

∴\therefore Molecular wt of virus = wt. of N A particle =

1.54×10−236.02×10−2×6.02×10−23  g/mol{{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}} \times 6.02 \times 10^{-23} \,\,g/mol

= 15400 g/mol = 15.4 kg/mol

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