Some Basic Concepts of Chemistry — Page 3 | NEET Chemistry

Q21

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H 2 SO 4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

A 1.4

B 3.0

C 2.8

D 4.4

Correct Answer

Option C

Solution

<table class=tg> <thead> <tr> <th class=tg-bzci></th> <th class=tg-bzci>HCOOH</th> <th class=tg-bzci>

\overset{{{H_2}S{O_4}}}\longrightarrow

</th> <th class=tg-bzci>H<sub>2</sub>O</th> <th class=tg-bzci>+</th> <th class=tg-bzci>CO</th> </tr> </thead> <tbody> <tr> <td class=tg-bzci>Initial</td> <td class=tg-bzci>

{{2.3} \over {46}} = {1 \over {20}}

</td> <td class=tg-bzci></td> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci>0</td> </tr> <tr> <td class=tg-bzci>Final</td> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci>

{1 \over {20}}

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

{1 \over {20}}

</td> </tr> </tbody> </table> <table class=tg> <thead> <tr> <th class=tg-bzci></th> <th class=tg-bzci>H<sub>2</sub>C<sub>2</sub>O<sub>4</sub></th> <th class=tg-bzci>

\overset{{{H_2}S{O_4}}}\longrightarrow

</th> <th class=tg-bzci>H<sub>2</sub>O</th> <th class=tg-bzci>+</th> <th class=tg-bzci>CO</th> <th class=tg-60hs>+</th> <th class=tg-60hs>CO<sub>2</sub></th> </tr> </thead> <tbody> <tr> <td class=tg-bzci>Initial</td> <td class=tg-bzci>

{{4.5} \over {90}} = {1 \over {20}}

</td> <td class=tg-bzci></td> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci>0</td> <td class=tg-60hs></td> <td class=tg-60hs>0</td> </tr> <tr> <td class=tg-bzci>Final</td> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci>

{1 \over {20}}

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

{1 \over {20}}

</td> <td class=tg-60hs></td> <td class=tg-60hs>

{1 \over {20}}

</td> </tr> </tbody> </table> <br><br>KOH will absorb CO<sub>2</sub> and conc.

H<sub>2</sub>SO<sub>4</sub> will absorb obtain water so in final solution only CO left. <br><br>Moles of CO formed from both reactions, <br>=

{1 \over {20}}

+

{1 \over {20}}

=

{1 \over {10}}

<br><br> $\therefore$ Mass of CO = moles × molar mass =

{1 \over {10}} \times 28

= 2.8 g

Q22

In which case is number of molecules of water maximum?

A 18 mL of water

B 0.18 g of water

C 0.00224 L of water vapours at 1 atm and 273 K

D 10 –3 mol of water

Correct Answer

Option A

Solution

(a) 18 mL water : Since, density of H 2 O = 1 g/mL Mass of water = V × d = 18 × 1 = 18 g No. of molecules of H 2 O =

{{18} \over {18}} \times

N A = 1 $\times$ N A (b) 0.18 g of water : No. of molecules of H 2 O =

{{0.18} \over {18}} \times

N A = 0.01 $\times$ N A (c) Volume of H 2 O(g) STP = 0.00224 L No. of molecules of H 2 O =

{V \over {22.4}}

$\times$ N A =

{{0.00224} \over {22.4}}

$\times$ N A = 0.0001 $\times$ N A (d) No of Molecules of water = mole × N A = 10 –3 N A

Q23

Which of the following is dependent on temperature?

A Molarity

B Mole fraction

C Weight percentage

D Molality

Correct Answer

Option A

Solution

Molarity = Number of moles of solute Volume of solution (in L) Molality = Number of moles of solute Mass of solvent (in kg) Mole fraction = Number of moles of component Total number of moles of all components Weight percentage = Weight of a component Total weight of solution $\times$ 100 KEY Concept : Those units which are volume related will be affected by the change in temperature.

The definition of molarity is - Number of moles of solute present in one litre of solution.

From definition you can see it is depends on volume which increases with increasing temperature and decreases with decreasing temperature.

All the other three options (Molality, Mole fraction, Weight fraction of solute) are mass related units and temperature has no effect on mass.

Q24

Suppose the elements X and Y combine to form two compounds XY 2 and X 3 Y 2 . When 0.1 mole of XY 2 weights 10 g and 0.05 mole of X 3 Y 2 weighs 9 g, the atomic weights of X and Y are

A 40, 30

B 60, 40

C 20, 30

D 30, 20

Correct Answer

Option A

Solution

Assume the atomic weight of element X is x and element Y is y. For XY 2 ,

n = {w \over {Mol.\,\,wt.}}

0.1 = {{10} \over {x + 2y}}

\Rightarrow x + 2y = {{10} \over {0.1}} = 100

...(i) For X 3 Y 2 ,

n = {w \over {Mol.\,\,wt.}}

0.05 = {9 \over {3x + 2y}}

\Rightarrow 3x + 2y = {9 \over {0.05}} = 180

...(ii) On solving equations (i) and (ii), we get y = 30 g mol -1 Put the value of y in equation. (i) x + 2(30) = 100 $\Rightarrow$ x = 100 - 60 = 40 g mol -1

Q25

What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO 3 is mixed with 50 mL of 5.8% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5

A 3.5 g

B 7 g

C 14 g

D 28 g

Correct Answer

Option B

Solution

50 ml of 16.9% solution of AgNO<sub>3</sub><br><br>

\left( {{{16.9} \over {100}} \times 50} \right)

= 8.45 g of AgNO<sub>3</sub><br><br> n<sub>mole</sub> =

{{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}

<br><br> =

\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles

<br><br> 50ml of 5.8% solution of NaCl contain<br><br> NaCl =

\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g

<br><br> n<sub>NaCl</sub> =

{{2.9g} \over {(23 + 35.5)g/mol}}

<br><br> = 0.0495 moles<br><br> <table class=tg> <tbody><tr> <th class=tg-baqh>AgNO<sub>3</sub></th> <th class=tg-baqh>+</th> <th class=tg-baqh>NaCl</th> <th class=tg-baqh> $\Rightarrow$ </th> <th class=tg-baqh>AgCl</th> <th class=tg-baqh>+</th> <th class=tg-baqh>Na<sup> $\oplus$ </sup></th> <th class=tg-baqh>+</th> <th class=tg-baqh>Cl<sup>

\ominus

</sup></th> </tr> <tr> <td class=tg-baqh>1 mole</td> <td class=tg-baqh></td> <td class=tg-baqh>1 mole</td> <td class=tg-baqh></td> <td class=tg-baqh>1 mole</td> <td class=tg-baqh></td> <td class=tg-baqh></td> <td class=tg-baqh></td> <td class=tg-baqh></td> </tr> <tr> <td class=tg-0lax>0.049 mole</td> <td class=tg-0lax></td> <td class=tg-0lax>0.049 mole</td> <td class=tg-0lax></td> <td class=tg-0lax>0.049 mole of AgCl</td> <td class=tg-0lax></td> <td class=tg-0lax></td> <td class=tg-0lax></td> <td class=tg-0lax></td> </tr> </tbody></table> <br> n =

{w \over M}

<br><br> $\Rightarrow$ w = (n<sub>AgCl</sub>) $\times$ Molecular mass<br><br> = (0.049) $\times$ (107.8 + 35.5) = 7.02 g

Q26

What is the mole fraction of the solute in a 1.00 m aqueous solution?

A 1.770

B 0.0354

C 0.0177

D 0.177

Correct Answer

Option C

Solution

Number of moles of water in 1000 g =

{{1000} \over {18}}

= 55.56 mol Thus, mole fraction of solute =

{1 \over {1 + 55.56}}

= 0.0177

Q27

The number of water molecules is maximum in

A 1.8 gram of water

B 18 gram of water

C 18 moles of water

D 18 molecules of water

Correct Answer

Option C

Solution

1.8 g of water =

{{6.023 \times {{10}^{23}}} \over {18}} \times 1.8

= 6.023 $\times$ 10 22 molecules $\therefore$ 18 g of water = 6.023 $\times$ 10 23 molecules = 1 mole of water $\therefore$ 18 moles of water = 18 $\times$ 6.023 $\times$ 10 23 molecules

Q28

If Avogadro number N A , is changed from 6.022

\times

10 23 mol

-

1 to 6.022

\times

10 20 mol

-

1 , this would change

A the mass of one mole of carbon

B the ratio of chemical species to each other in a balanced equation

C the ratio of elements to each other in a compound

D the definition of mass in units of grams.

Correct Answer

Option A

Solution

We know mass of 1 mol (6.022 $\times$ 10 23 ) atoms of carbon = 12 g If Avogadro number is changed to 6.022 $\times$ 10 20 then mass of 1 mol of cabon is =

{{12 \times 6.022 \times {{10}^{20}}} \over {6.022 \times {{10}^{23}}}}

=

12 \times {10^{ - 3}}\,g

$\therefore$ It would change the mass of one mole of carbon.

Q29

A mixture of gases contains H 2 and O 2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture ?

A 16 : 1

B 2 : 1

C 1 : 4

D 4 : 1

Correct Answer

Option D

Solution

Ratio of weight of gases = w H 2 : w O 2 = 1 : 4 $\therefore$ Number of moles of H 2 =

1 \over 2

$\therefore$ Number of moles of O 2 =

4 \over 32

Ratio of moles of gases = n H 2 : n O 2 =

1 \over 2

:

4 \over 32

=

1 \over 2

$\times$

32 \over 4

= 4 : 1

Q30

When 22.4 liters of H 2 (g) is mixed with 11.2 litres of Cl 2 (g) , each at S.T.P. the moles of HCl (g) formed is equal to

A 1 mol of HCl (g)

B 2 mol of HCl (g)

C 0.5 mol of HCl (g)

D 1.5 mol of HCl (g)

Correct Answer

Option A

Solution

1 mole = 22.4 L at S.T.P<br><br> n<sub>H<sub>2</sub></sub> =

{{22.4} \over {22.4}}

= 1 mole<br><br> n<sub>Cl<sub>2</sub></sub> =

{{11.2} \over {22.4}}

= 0.5 mole<br><br> <table class=tg> <tbody><tr> <th class=tg-s6z2></th> <th class=tg-s6z2>H<sub>2</sub></th> <th class=tg-s6z2>+</th> <th class=tg-s6z2>Cl<sub>2</sub></th> <th class=tg-s6z2> $\Rightarrow$ </th> <th class=tg-s6z2>2HCl<sub>(g)</sub></th> </tr> <tr> <td class=tg-s6z2>Initial</td> <td class=tg-s6z2>1 mol</td> <td class=tg-s6z2></td> <td class=tg-s6z2>0.5 mol</td> <td class=tg-s6z2></td> <td class=tg-s6z2>0</td> </tr> <tr> <td class=tg-s6z2>Final</td> <td class=tg-s6z2>( 1 - 0.5) <br> = 0.5 mol</td> <td class=tg-s6z2></td> <td class=tg-s6z2>(0.5 - 0.5) <br> = 0 mol</td> <td class=tg-s6z2></td> <td class=tg-s6z2>2 x 0.5 <br> 1 mol</td> </tr> </tbody></table><br> Here Cl<sub>2</sub> is limiting reagent.

So 1 mole of HCl<sub>(g)</sub> is formed.