Some Basic Concepts of Chemistry — Page 2 | JEE Chemistry

Q11

When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of

A

?

A

\mathrm{C}_{9} \mathrm{H}_{8}

B

\mathrm{C}_{5} \mathrm{H}_{8}

C

\mathrm{C}_{11} \mathrm{H}_{4}

D

\mathrm{C}_{11} \mathrm{H}_{8}

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\\\ & \frac{\mathrm{y}}{2}=4 \Rightarrow \mathrm{y}=8 \\\\ & \mathrm{x}+\frac{8}{4}=11 \\\\ & \Rightarrow \mathrm{x}=9 \\\\ & \therefore \text{ Hydrocarbon will be }=\mathrm{C}_9 \mathrm{H}_8 \end{aligned}

Q12

A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is: (Atomic mass : Cl = 35.5 amu, H = 1 amu)

A 10.2 M

B 12.5 M

C 14.0 M

D 18.2 M

Correct Answer

Option C

Solution

35% HCl by mass means in 100 gm HCl solution 35 gm HCl present. Now, volume of 100 gm HCl solution

= {{100} \over {1.46}}

ml

= {{{{100} \over {1.46}}} \over {1000}}

l Moles of HCl

= {{35} \over {36.5}}

Now, molarity

= {{moles\,of\,solute} \over {volume\,of\,solution\,(in\,L)}}

= {{{{35} \over {36.5}}} \over {{{{{100} \over {1.46}}} \over {1000}}}} = 14

Q13

A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution ?

A 0.190

B 0.086

C 0.050

D 0.100

Correct Answer

Option B

Solution

The formula between Mole fraction and Molality is

m = {{{X_{solute}}} \over {{X_{solvent}}}} \times {{1000} \over {{M_{solvent}}}}

$

m

= molality, Xsolute = Mole fraction of solute, Xsolvent = Mole fraction of solvent, Msolvent = Molar mass of solvent Here solute is methyl alcohol(CH3OH) and solvent is water(H2O).

Here water is solvent because question says solution is aqueous.

Assume mole fraction of methyl alcohol = X So mole fraction of water = 1 - X and given molality (m) = 5.2 Msolvent = 18 (as molar mass of H2O = 18) $\therefore$ 5.2 =

{X \over {1 - X}} \times {{1000} \over {18}}

by solving this we get, X = 0.086

Q14

The minimum amount of O2(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

A 4Fe(s) + 3O2(g)

\to

2Fe2O3(s)

B P4(s) + 5O2(g)

\to

P4O10(s)

C C3H8(g) + 5O2(g)

\to

3CO2(g) + 4H2O(l)

D 2Mg(s) + O2(g)

\to

2MgO(s)

Correct Answer

Option A

Solution

(a) 4Fe(s) + 3O2(g) $\to$ 2Fe2O3(s)

{{Moles\,of\,{O_2}} \over 3} = {{Moles\,of\,{Fe}} \over 4}

$\Rightarrow$ Moles of O2 =

{3 \over 4}

$\times$

{1 \over {56}}

moles =

{1 \over {24.8}}

moles =

{3 \over {224}} \times 32

g of O2 = 0.43 g of O2 (b) P4(s) + 5O2(g) $\to$ P4O10(s)

{{Moles\,of\,{O_2}} \over 5} = {{Moles\,of\,{P_4}} \over 1}

$\Rightarrow$ Moles of O2 = 5 $\times$

{1 \over {124}}

moles =

{1 \over {24.8}}

moles =

{1 \over {24.8}} \times 32

g of O2 = 1.3 g of O2 (c) C3H8(g) + 5O2(g) $\to$ 3CO2(g) + 4H2O(l)

{{Moles\,of\,{O_2}} \over 5} = {{Moles\,of\,{C_3}{H_8}} \over 1}

$\Rightarrow$ Moles of O2 = 5 $\times$

{1 \over {44}}

moles =

{1 \over {24.8}}

moles =

{1 \over {8.8}} \times 32

g of O2 = 3.6 g of O2 (d) 2Mg(s) + O2(g) $\to$ 2MgO(s)

{{Moles\,of\,{O_2}} \over 1} = {{Moles\,of\,{Mg}} \over 2}

$\Rightarrow$ Moles of O2 =

{1 \over 2}

$\times$

{1 \over {24}}

moles =

{1 \over {48}}

moles =

{1 \over {48}} \times 32

g of O2 = 0.66 g of O2

Q15

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is

A 40 mL

B 20 mL

C 10 mL

D 60 mL

Correct Answer

Option A

Solution

H3PO3 + 2KOH $\to$ K2HPO3 + 2H2O Let the volume of KOH solution is = V mL =

{V \over {1000}}

litre Volume of H3PO3 = 20 mL =

{{20} \over {1000}}

litre No of moles of H3PO3 present =

{{20} \over {1000}} \times 0.1

No of moles of KOH present =

{V \over {1000}} \times 0.1

Stoichiometric coefficient of H3PO3 = 1 Stoichiometric coefficient of KOH = 2 $\therefore$

{{{{20} \over {1000}} \times 0.1} \over 1}

=

{{{V \over {1000}} \times 0.1} \over 2}

$\Rightarrow$ V = 40 mL

Q16

Density of a 2.05M solution of acetic acid in water is 1.02 g/mL The molality of the solution is

A 2.28 mol kg-1

B 0.44 mol kg-1

C 1.14 mol kg-1

D 3.28 mol kg-1

Correct Answer

Option A

Solution

2.05M solution of acetic acid in water means in 1 litre solution 2.05 moles of CH3COOH present.

Density of solution = 1.02 g/ml (Given) Assume the volume of solution = 1 litre = 1000 ml $\therefore$ Mass of solution = 1000 $\times$ 1.02 = 1020 gm Molar mass of CH3COOH = 60 So Mass of CH3COOH (msolute) = 2.05 $\times$ 60 = 123 $\therefore$ Mass of solvent (msolvent) = 1020 - 123 = 897 gm = 0.897 kg Formula of molality (m) =

{{no\,of\,moles\,of\,solute} \over {weight\,of\,solvent\,in\,kg}}

$\therefore$ m =

{{2.05} \over {0.897}}

= 2.28 Using Formula : Molality (m) =

{{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}

Here M = molarity, Msolute = molecular mass of solute, d = density of solution $\therefore$ m =

{{1000 \times 2.05} \over {1000 \times 1.02 - 2.05 \times 60}}

= 2.28

Q17

In the reaction 2Al(s) + 6HCl(aq)

\to

2Al3+ (aq) + 6Cl-(aq) + 3H2(g)

A 11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed

B 6 L HCl(aq) is consumed for every 3L H2(g) produced

C 33.6 L H2(g) is produced regardless of temperature and pressure for every mole that reacts

D 67.2 H2(g) at STP is produced for every mole Al that reacts

Correct Answer

Option A

Solution

Option A : 6 moles of HCl(aq) produces = 3 moles of H2(g) $\therefore$ 1 mole of HCl(aq) produces =

{3 \over 6}

moles of H2(g) =

{1 \over 2}

moles of H2(g) From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L. $\therefore$

{1 \over 2}

moles of H2(g) occupies =

{1 \over 2} \times 22.4

= 11.2 L $\therefore$ Option (A) is correct.

Option B : According to avogadro Hypothesis, volume is directly proportional to the no of moles of gases.

Only for gases, no of mole and volume relationship is possible i.e, for two gases if no of moles are same then their volume is also same.

But here HCL is in aqueous form and H2 is in gaseous form so we can't find volume of HCL(aq) using no of moles of HCL(aq).

$\therefore$ Option (B) is wrong.

Option C : 2 moles of Al(s) produces = 3 moles of H2(g) $\therefore$ 1 mole of Al(s) produces =

{3 \over 2}

moles of H2(g) From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L. $\therefore$

{3 \over 2}

moles of H2(g) occupies =

{3 \over 2} \times 22.4

= 33.6 L 6 moles of HCl(aq) produces = 3 moles of H2(g) $\therefore$ 1 mole of HCl(aq) produces =

{3 \over 6}

moles of H2(g) =

{1 \over 2}

moles of H2(g) $\therefore$

{1 \over 2}

moles of H2(g) occupies =

{1 \over 2} \times 22.4

= 11.2 L So from every mole of HCl(aq) that reacts only 11.2 L H2(g) produces.

Form every mole of Al(s) produces 33.6 L H2(g) but from every mole of HCl(aq) that reacts only 11.2 L H2(g) produces not 33.6 L.

$\therefore$ Option (C) is wrong.

Option D : 2 moles of Al(s) produces = 3 moles of H2(g) $\therefore$ 1 mole of Al(s) produces =

{3 \over 2}

moles of H2(g) From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L. $\therefore$

{3 \over 2}

moles of H2(g) occupies =

{3 \over 2} \times 22.4

= 33.6 L $\therefore$ Option (D) is wrong.

Q18

The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol-1) by mass will be

A 1.45

B 1.64

C 1.88

D 1.22

Correct Answer

Option D

Solution

Important formula of Molarity(M) when % w/w is given M =

{{10 \times \% w/w \times d} \over {{M_{solute}}}}

Here M = 3.6, % w/w = 29, d = density, Msolute = 98 $\therefore$ 3.6 =

{{10 \times 29 \times d} \over {98}}

\Rightarrow d = 1.22

Q19

The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is:

A 37.5 kg

B 7.5 kg

C 10 kg

D 15 kg

Correct Answer

Option B

Solution

Given that weight of human adult is = 75 kg Among those 75 kg, 10% is Hydrogen(1H). $\therefore$ Mass of 1H =

75 \times {{10} \over {100}}

= 7.5 kg Now when every 1H atom is replaced by 2H atom then weight of every atom is become double.

So total weight of 2H becomes = 2 $\times$ 7.5 = 15 kg.

So the weight gain by the person = 15 - 7.5 = 7.5 kg

Q20

An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are : (Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023

\times

1023 mol-1)

A 6.023

\times

1020

B 6.023

\times

109

C 6.023

\times

1021

D 6.023

\times

1023

Correct Answer

Option A

Solution

% of Cl = 3.55

\therefore\,\,\,\,

In 100 g chlorohydrocarbon 3.55 gm Cl present. In 1 gm chlorohydrocarbon Cl present =

{{3.55} \over {100}}

= 0.0355 gm

\therefore\,\,\,\,

No of Moles of Cl =

{{0.0355} \over {35.5}}

= 0.001 mole

\therefore\,\,\,\,

no of Cl atoms = 0.001 $\times$ 6.023 $\times$ 103 = 6.023 $\times$ 1020