Some Basic Concepts of Chemistry — Page 10 | NEET Chemistry

Q91

Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is, Given : Atomic masses of C, H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 g mol

-

1.

A C4H6N2

B C2H3N

C C5H7N

D C10H14N2

Correct Answer

Option D

Solution

Mole ratio of H, C and N

= {{8.7} \over 1}:{{74} \over {12}}:{{17.\,3} \over {14}}

= 8\,.7:6.167:1.23

= {{8.7} \over {1.23}}:{{6.167} \over {1.23}}:{{1.23} \over {1.23}}

= 7:5:1

$\therefore$ Emperical formula =

{{C_5}{H_7}N}

$\therefore$ Molecular formula

= {\left( {{C_5}{H_7}N} \right)_n}

Given molecular mass = 162 Molecular mass of

{\left( {{C_5}{H_7}N} \right)_n}

= (5 \times 12 + 7 \times 1 + 14) \times n

= (81) \times n

$\therefore$

81 \times n = 162

\Rightarrow n = 2

$\therefore$ Molecular formula = C10H14N2

Q92

What would be the molality of 20% (mass/ mass) aqueous solution of KI? (molar mass of KI = 166 g mol–1)

A 1.51

B 1.35

C 1.08

D 1.48

Correct Answer

Option A

Solution

20% (mass/ mass) aqueous solution means, 100 g solution contains 20 g KI $\therefore$ Mass of solvent = 100 – 20 = 80 g So, 20 g of KI present in 80 g of solvent.

$\therefore$

{{20} \over {166}}

moles of KI present in 80 g of solvent. We know, Molality (m) =

{{no\,of\,moles\,of\,solute} \over {wt(in\,kg)\,of\,solvent}}

$\therefore$ m =

{{{{20} \over {166}}} \over {{{80} \over {1000}}}}

= 1.51

Q93

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?

A 50 mL

B 12.5 mL

C 25 mL

D 75 mL

Correct Answer

Option C

Solution

2HCl(aq) + Na2CO3(aq) $\to$ H2CO3 + NaCl moles of HCl 2 = moles of Na2CO3 1 $\Rightarrow$

{{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}

$\Rightarrow$ Molarity of HCl (M) =

{6 \over {25}}M

HCl(aq) + NaOH(aq) $\to$ NaCl + H2O moles of HCl 1 = moles of NaOH 1

{{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}

$\Rightarrow$ V = 25 ml

Q94

A sample of

NaCl{O_3}

is converted by heat to

NaCl

with a loss of

0.16

g

of oxygen. The residue is dissolved in water and precipitated as

AgCl.

The mass of

AgCl

(in

g

) obtained will be : (Given : Molar mass of

AgCl=143.5

g

mo{l^{ - 1}}

)

A

0.35

B

0.41

C

0.48

D

0.54

Correct Answer

Option C

Solution

NaClO3

\overset{\Delta}\longrightarrow

2NaCl + 3O2 Here O2 produced = 0.16 g

\therefore\,\,\,\,

No of moles of O2 =

{{0.16} \over {32}}

= 5 $\times$ 10 $-$ 3 Let no of moles of NaCl

\therefore\,\,\,\,

{{{}^nNaCl} \over 2}

=

{{^n{O_2}} \over 3}

$\Rightarrow$

\,\,\,\,

^nNaCl

=

{2 \over 3}

$\times$ 5 $\times$ 10 $-$ 3 =

{1 \over {300}}

NaCl + Ag+ $\to$ AgCl + Na+ From here, you can see no of moles in NaCl = no of moles in AgCl

\therefore\,\,\,\,

no of moles in AgCl =

{1 \over {300}}

\therefore\,\,\,\,

Mass of AgCl = 143.5 $\times$

{1 \over {300}}

= 0.48 g

Q95

10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is :

A 0.2 M NaCl solution

B 20 M HCl solution

C Neutral solution

D 10 M HCl solution

Correct Answer

Option B

Solution

When $10 \mathrm{ml}, 2 \mathrm{M} \mathrm{NaOH}$ solution is added to 20 ml of 1 M HCl solution : $\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$

\begin{array}{rlrl} \text{ Initial : MV } & =2 \times 0.1 & & M V=1 \times 0.2 \\ & =0.2 \text{ mole } & & =0.2 \mathrm{~mole} \\ \text{ Final } & 0 & 0 \end{array}

$\therefore$ Resulting solution becomes neutral.

Now when 10 mol of above solution is poured into a flask containing 2 mole HCl and made solution 100 ml will distilled water.

Molarity of $\mathrm{HCl}=\dfrac{2}{100} \times 1000=20$

Q96

Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol-1) is

A twice that in 60 g carbon

B 6.023

\times

1022

C half that in 8 g He

D 558.5

\times

6.023

\times

1023

Correct Answer

Option A

Solution

Number of moles of Fe in 558.5 gm of Fe is $=$

{{558.5} \over {55.85}}

$=$ 10 $\therefore$ Number moles of Fe in 558.5 gm Fe is = 10 Since here Fe is monoatomic then no. of atoms in 10 moles Fe is $\Rightarrow$ 10 $\times$ 1 $\times$ NA $\Rightarrow$ 10 NA For Option (A) Atomic Mass of Carbon is 12 g then no. of moles in 60 g Carbon is $\Rightarrow$

{{60} \over {12}} = 5

Here Carbon is also monoatomic then no. of atoms in 5 moles of Carbon $\Rightarrow$ 5 $\times$ 1 $\times$ NA $\Rightarrow$ 5 NA Now we can clearly see 10 is twice of 5, so option (A) is correct.

Q97

For a reaction, N2(g) + 3H2(g)

\to

2NH3(g) ; identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.

A 56g of N2 + 10g of H2

B 14g of N2 + 4g of H2

C 28g of N2 + 6g of H2

D 35g of N2 + 8g of H2

Correct Answer

Option A

Solution

Here you have to check every options to find answer.

N2(g) + 3H2(g) $\to$ 2NH3(g) For dihydrogen (H2) to become a limiting reagent,

{{moles\,of\,{H_2}} \over {stoichiometric\,coefficient\,of\,{H_2}}}

should be less than

{{moles\,of\,{N_2}} \over {stoichiometric\,coefficient\,of\,{N_2}}}

. Here in 56g of N2 + 10g of H2, Moles of N2 =

{{56} \over {28}}

= 2 and Moles of H2 =

{{10} \over 2}

= 5 Now by dividing stoichiometric coefficient we get, For N2 =

{2 \over 1}

= 2 For H2 =

{5 \over 3}

= 1.67 $\therefore$ Here H2 is the limiting reagent.

Q98

Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be

A 7.01%

B 4.08%

C 6.05%

D 5.08%

Correct Answer

Option B

Solution

Given metal oxide = M0.98O We know oxidation number of O = (-2) Now assume oxidaton no of M =

x

$\therefore$ 0.98

x

+ 1 $\times$ (-2) = 0

\Rightarrow x =

{{200} \over {98}}

This represent the charge in one atom of M. As you can see the charge of M is in the range

2 < {{200} \over {98}} < 3

.

So we can say in M mixture of M+2 and M+3 present.

Assume total no of atoms present in M is 100.

Let M+3 present in M = y atoms So M+2 present in M = (100 - y) atoms In 1 atom of M+3 charge present = +3 So in y atoms of M+3 charge present = +3y Similarly in 1 atom of M+2 charge present = +2 So in (100 - y) atoms of M+2 charge present = +2(100 - y) $\therefore$ Total charge = 200 - 2y + 3y = 200 + y In 100 atoms of M total charge = 200 + y So in 1 atoms of M total charge =

{{200 + y} \over {100}}

Earlier we found that charge in one atom of M is =

{{200} \over {98}}

So we can write,

{{200 + y} \over {100}}

=

{{200} \over {98}}

\Rightarrow y = 4.08

So Option (B) is correct.

Q99

In Carius method of estimation of halogen,

0.45 \mathrm{~g}

of an organic compound gave

0.36 \mathrm{~g}

of

\mathrm{AgBr}

. Find out the percentage of bromine in the compound. (Molar masses :

\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}

)

A 34.04%

B 40.04%

C 36.03%

D 38.04%

Correct Answer

Option A

Solution

Mass of organic compound $=0.45 \,\mathrm{gm}$ Mass of $\mathrm{AgBr}$ obtained $=0.36 \,\mathrm{gm}$ $\therefore$ Moles of $\mathrm{AgBr}=\dfrac{0.36}{188}$ $\therefore$ Mass of Bromine $=\dfrac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}$ $\therefore \%\, \mathrm{Br}$ in compound $=\dfrac{0.1532}{0.45} \times 100=34.04 \,\%$