Some Basic Concepts of Chemistry — Page 4 | NEET Chemistry

Q31

Equal masses of H 2 , O 2 and methane have been taken in a container of volume V at temperature 27 o C in identical conditions. The ratio of the volumes of gases H 2 : O 2 : methane would be

A 8 : 16 : 1

B 16 : 8 : 1

C 16 : 1 : 2

D 8 : 1 : 2

Correct Answer

Option C

Solution

According to Avogadro's hypothesis ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no of moles =

{{Mass} \over {Mol.\,\,mass}}

n H 2 =

{w \over 2}

; n O 2 =

{w \over 32}

; n CH 4 =

{w \over 16}

So, the ratio will be

{w \over 2}

:

{w \over 32}

:

{w \over 16}

or 16 : 1 : 2

Q32

1.0 g of magnesium is burnt with 0.56 g O 2 in a closed vessel, Which reactant is left in excess and how much ? (At. wt. Mg = 24, O = 16)

A Mg. 0.16 g

B O 2 , 0.16 g

C Mg, 0.44 g

D O 2 , 0.28 g

Correct Answer

Option A

Solution

n<sub>Mg</sub> =

1 \over 24

= 0.0416 moles<br><br> n<sub>O<sub>2</sub></sub> =

0.56 \over 32

= 0.0175 moles<br><br> <table class=tg> <tbody><tr> <th class=tg-uys7></th> <th class=tg-uys7>Mg</th> <th class=tg-uys7>+</th> <th class=tg-uys7>

{1\over2} O_2

</th> <th class=tg-uys7> $\Rightarrow$ </th> <th class=tg-uys7>MgO</th> </tr> <tr> <td class=tg-uys7>Initial</td> <td class=tg-uys7>0.0416 moles</td> <td class=tg-uys7></td> <td class=tg-uys7>0.0175 moles</td> <td class=tg-uys7></td> <td class=tg-uys7>0</td> </tr> <tr> <td class=tg-uys7>Final</td> <td class=tg-uys7>( 0.0416 - 2 x 0.0175) <br> = 0.0066 moles</td> <td class=tg-uys7></td> <td class=tg-uys7>0</td> <td class=tg-uys7></td> <td class=tg-uys7>2 x 0.0175</td> </tr> </tbody></table> <br> $\because$ Mass of Mg left in excess = 0.0066 $\times$ 24 = 0.16 g

Q33

In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO 3 , which of the following will be the formula of the chloriode (X stands for the symbol of the element other than chlorine)

A X 2 Cl 2

B XCl 2

C XCl 4

D X 2 Cl

Correct Answer

Option B

Solution

Milimoles of solution of chloride = 0.05 $\times$ 10 = 0.5 Millimoles of AgNO 3 solution = 10 $\times$ 0.1 = 1 So, the millimoles of AgNO 3 are double than the chloride solution.

$\therefore$ XCl 2 + 2AgNO 3 $\to$ 2AgCl + X(NO 3 ) 2

Q34

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO 3 ? The concentrated acid is 70% HNO 3 .

A 70.0 g conc. HNO 3

B 54.0 g conc. HNO 3

C 45.0 g conc. HNO 3

D 90.0 g conc. HNO 3

Correct Answer

Option C

Solution

2 = {m \over {63 \times 0.25}}

$\Rightarrow$ m = 2 × 63 × 0.25 = 31.5 g Now, if concentrated HNO 3 is 100% then it requires 31.5 g.

But the original solution of HNO 3 is 70% concentrated.

Hence, the mass of HNO 3 required to produce 2.0 M solution =

{{100} \over {75}} \times 31.5

= 45 g of HNO 3

Q35

6.02

\times

10 20 molecules of urea present in 100 mL of its solution. The concentration of solution is

A 0.001 M

B 0.1 M

C 0.02 M

D 0.01 M

Correct Answer

Option D

Solution

Moles of urea =

{{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}} = 0.001

Concentration of solution =

{{0.001} \over {100}} \times 1000

= 0.01 M

Q36

Which has the maximum number of molecules among the following ?

A 44 g CO 2

B 48 g O 3

C 8 g H 2

D 64 g SO 2

Correct Answer

Option C

Solution

8 g H 2 has 4 moles while the others has 1 mole each.

Q37

The number of atoms in 0.1 mol of a triatomic gas is (N A = 6.02

\times

10 23 mol

-

1 )

A 6.026

\times

10 22

B 1.806

\times

10 23

C 3.600

\times

10 23

D 1.800

\times

10 22

Correct Answer

Option B

Solution

No. of atoms = N A $\times$ No. of moles $\times$ 3 = 6.023 $\times$ 10 23 $\times$ 0.1 $\times$ 3 = 1.806 $\times$ 10 23

Q38

25.3 g of sodium carbonate, Na 2 CO 3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na + and carbonate ions, CO

{}_3^{2 - }

are respectively (Molar mass of Na 2 CO 3 = 106 g mol

-

1 )

A 0.955 M and 1.910 M

B 1.910 M and 0.955 M

C 1.90 M and 1.910 M

D 0.477 M and 0.477 M

Correct Answer

Option B

Solution

Given in the question that molar mass of Na 2 CO 3 = 106 g $\therefore$ Molarity of solution =

{{25.3 \times 1000} \over {106 \times 250}}

= 0.9547 M = 0.955 M Na 2 CO 3 $\to$ 2Na + +

CO_3^{2-}

[Na + ] = 2[Na 2 CO 3 ] = 2 $\times$ 0.955 = 1.910 M [

CO_3^{2-}

] = [Na 2 CO 3 ] = 0.955 M

Q39

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

A 3 mol

B 4 mol

C 1 mol

D 2 mol

Correct Answer

Option B

Solution

<table class=tg> <tbody><tr> <th class=tg-s6z2>H2</th> <th class=tg-s6z2>+</th> <th class=tg-s6z2>1/2O2</th> <th class=tg-s6z2> $\to$ </th> <th class=tg-s6z2>H2O</th> </tr> <tr> <td class=tg-s6z2>2 g</td> <td class=tg-s6z2></td> <td class=tg-s6z2>16 g</td> <td class=tg-s6z2></td> <td class=tg-s6z2>18 g</td> </tr> <tr> <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>1 mol</td> </tr> </tbody></table> 10 g of H<sub>2</sub> = 5 mol and 64 g og O<sub>2</sub> = 2 mol<br><br> $\therefore$ In this reaction, oxygen is the limiting reagent so amount of H<sub>2</sub>O produced depends on that of O<sub>2</sub><br><br> Since 0.5 mol of O<sub>2</sub> gives 1 mol H<sub>2</sub>O<br><br> $\therefore$ 2 mol of O<sub>2</sub> will give <b>4 mol</b> H<sub>2</sub>O

Q40

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl ?

A 0.011

B 0.029

C 0.044

D 0.333

Correct Answer

Option B

Solution

6.5 \over 224

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

3.2 \over 36.5

mol</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-baqh>0.029 mol</td> <td class=tg-baqh></td> <td class=tg-baqh>0.087 mol </td> <td class=tg-baqh></td> <td class=tg-baqh></td> <td class=tg-baqh></td> <td class=tg-baqh></td> </tr> </tbody></table> <br> Formation of moles of lead(II) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here.

So no of moles of PbCl<sub>2</sub> formed will be equal to the no of PbO (i.e 0.029)