Some Basic Concepts of Chemistry

C_xH_y + $\left( {x + {y \over 4}} \right)$O₂ $ \to $ xCO₂ + ${y \over 2}$H₂O

No of moles of CO₂ produced = ${{88} \over {44}}$ = 2 mol
No of moles of H₂O produced = ${9 \over {18}}$ = 0.5 mol

According to stoichiometric mole mole analysis :

${{moles\,of\,C{O_2}} \over x} = {{moles\,of\,{H_2}O} \over {{y \over 2}}}$
$ \Rightarrow $ ${2 \over x} = {{0.5} \over {{y \over 2}}}$
$ \Rightarrow $ ${x \over y} = {2 \over 1}$

So, ${{no\,of\,atoms\,of\,'C'\,in\,{C_x}{H_y}} \over {no\,of\,atoms\,of\,'H'\,in\,{C_x}{H_y}}} = {2 \over 1}$

$ \therefore $ ${{Mass\,of\,'C'\,in\,{C_x}{H_y}} \over {Mass\,of\,'H'\,in\,{C_x}{H_y}}} = {{2 \times 12} \over {1 \times 1}} = {{24} \over 1}$

KMnO4	$\buildrel {513\,\,K} \over \longrightarrow $	K2MnO4	+	MnO2	+	O2
(X)		(Y)

MnO2	+	NaCl	+	conc H2SO4	$ \to $	MnSO4	+	NaHSO4	+	H2O	+	Cl2
												(Z)

We know,

molality(m) = ${{{x_{solute}}} \over {{x_{solvent}}}} \times {{1000} \over {molar\,weight\,of\,solvent}}$

Here solvent is H₂O

x_solvent = 0.8

$ \therefore $ x_solute = 0.2

$ \therefore $ Molality = ${{0.2} \over {0.8}} \times {{1000} \over {18}}$ = 13.88

Molar weigh of 'A' = M_A

Molar weigh of 'B' = M_B

Molar mass of AB₂ = M_A + 2M_B

Molar mass of A₂B₂ = 2(M_A + M_B)

5 mol AB₂ weighs 125 g

$ \therefore $ AB₂ = 25 g/mol

10 mol A₂B₂ weighs 300 g

$ \therefore $ A₂B₂ = 30 g/mol

So, M_A + 2M_B = 25 ..............(1)

and 2(M_A + M_B) = 30 ..............(2)

By solving (1) and (2),

M_A = 5 gm = 5 $ \times $ 10^-3 kg

M_B = 10 gm = 10 $ \times $ 10^-3 kg

(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

C_xH_y + $\left( {x + {y \over 4}} \right)$O₂ $ \to $ xCO₂ + ${y \over 2}$H₂O

According to the above equation,

1 mL of hydrocarbon = x mL of CO₂ is produced

$ \therefore $ 10 mL of hydrocarbon = 10x mL of CO₂ is produced.

According to question,

From 10 mL of hydrocarbon 40 mL of CO₂ is produced.

$ \therefore $ 10x = 40

$ \Rightarrow $ x = 4

Also from the above equation,

1 mL of C_xH_y react with $\left( {x + {y \over 4}} \right)$ mL of O₂.

$ \therefore $ 10 mL of C_xH_y react with 10$\left( {x + {y \over 4}} \right)$ mL of O₂.

According to question,

10$\left( {x + {y \over 4}} \right)$ = 55

$ \Rightarrow $ $40 + {{10y} \over 4} = 55$

$ \Rightarrow $ y = 6

$ \therefore $ Hydrocarbon is C₄H₆

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

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.

In CH₄, there are 5 atoms (1C + 4H).

$ \therefore $ % composition of C by mole in CH₄

= ${1 \over 5}$ $ \times $ 100 = 20 %

8 gm of NaOH   =   ${8 \over {40}}$   =   0.2 mol of NaOH

18 gm of H₂O   =   ${18 \over {18}}$   =   1 mol of H₂O

$ \therefore $  Total mole   =   1 + 0.2   =   1.2 mol

$ \therefore $  Mole fraction of NaOH   =   ${{0.2} \over {1.2}}$   =   0.167

We know,

Molality   =   ${{Moles\,\,of\,\,solute} \over {Weight\,of\,solvent}} \times $ 1000

     =  ${{0.2} \over {18}} \times $ 1000

     =  11.11

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

Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1

Thus, the correct formula is C₆H₈N₂. .

2NaHCO₃ + (COOH)₂ $ \to $ (COONa)₂ + 2H₂O + 2CO₂

$ \therefore $ 1 mole of CO₂ is produced by 1 mole of NaHCO₃.

Given, volume of CO₂ produced = 0.25 ml

25 L of CO₂ contains 1 mol

$ \therefore $ 0.25 ml of CO₂ contains = ${1 \over {25 \times {{10}^3}}} \times 0.25$ moles

= 10^-5 moles

$ \therefore $ Moles of NaHCO₃ = 10^-5 moles

Mass of NaHCO₃ = 10^-5 $ \times $ 84 g

$ \therefore $ % of NaHCO₃ in a tablet

= ${{84 \times {{10}^{ - 5}}} \over {10 \times {{10}^{ - 3}}}} \times 100$ = 8.4 %

Molarity = ${{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}$

0.1 = ${{wt./342} \over 2}$

wt (C₁₂H₂₂O₁₁) = 68.4 gram

moles of C₅₇H₁₁₀O₆(s) = ${{445} \over {890}}$ = 0.5 moles

2C₅₇H₁₁₀O₆(s) + 163 O₂(g) $ \to $ 114 CO₂(g) + 110 H₂O(l)

n_H2O = ${{110} \over 4}$ = ${{55} \over 2}$

m_H2O = ${{55} \over 2}$ $ \times $ 18

= 495 gm

Molality of Na⁺ = ${{Moles\,\,of\,\,N{a^ + }} \over {mass\,\,of\,\,{H_2}O\,\,in\,\,kg}}$

Moles of Na⁺ = ${{92} \over {23}}$ = 4

Mass of H₂O = 1 kg

$ \therefore $   Molality = ${4 \over 1}$ = 4

%    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

$\therefore\,\,\,$ The ratio of no of atoms of C and H in one molecule of C_x H_y O_z = 1 : 2

$\therefore\,\,\,$ y = 2x

In one molecule of C_x H_y O_z compound contain z atoms of oxygen.

According to the question, no of atoms of oxygen required to burn C_xH_y completely should be twice of z atoms of oxygen.

C_xH_y + (x + ${y \over 4}$) O₂ $ \to $ HCO₂ + ${y \over 2}$ H₂O

No. of O₂ molecules required = (x + ${y \over 4}$)

$\therefore\,\,\,$ No. of O atoms required = 2 (x + ${y \over 4}$)

According to question,

$2\left( {x + {y \over 4}} \right) = 2z$

$ \Rightarrow \,\,\,2\left( {x + {{2x} \over 4}} \right) = 2z\,\,\,$ [ Putting y = 2x ]

$ \Rightarrow \,\,\,\,{{3x} \over 2} = z$

$\therefore\,\,\,$ x : y : z

= x : 2x : ${{3x} \over 2}$

= 2 : 4 : 3

$\therefore\,\,\,$ Empirical formula = C₂H₄O₃

(a)   Molar mass of (NH₄)₂ Cr₂O₇ = 252 g/mol.

252g of (NH₄)₂ Cr₂ O₇ produce 28g mole of N₂

$\therefore\,\,\,$ 1 g of (NH₄)₂ Cr₂ O₇ Produce = ${{28} \over {252}} = 0.111$ g N₂

(b)    Molar mass of NH₄ NO₃ = 80 g/mol

2 $ \times $ 80 g of NH₄ NO₃ produce 28 $ \times $ 2g of N₂

$\therefore\,\,\,$ 1 g of NH₄ NO₃ produce = ${{28 \times 2} \over {2 \times 80}}$ = 0.35 g N₂

(c)    Molar mass of Ba(N₃)₂ = 221 g/mol

221 g of Ba(N₃)₂ produce 3 $ \times $ 28 g of N₂

$\therefore\,\,\,$ 1 g of Ba(N₃)₂ produce = ${{3 \times 28} \over {221}}$ = 0.38 g of N₂

(d)    Molar mass of NH₃ = 17 g/mol

17 $ \times $ 2 NH₃ produce 28 g of N₂

$\therefore\,\,\,$ 1 g of NH₃ produce = ${{28} \over {17 \times 2}}$ = 0.823 g of N₂

NaClO₃ $\buildrel \Delta \over \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

Let the volume of a monoprotic acid solution in mL be V

${{V \times 45} \over {100}} + {{(800 - V)20} \over {100}} = {{800 \times 29.875} \over {100}}$

${{9V} \over {20}} + 160 - {V \over 5} = 239$

${{5V} \over {20}} = 79 \Rightarrow V = 316$ mL

3 NaOH (aq.) + FeCl₃(aq) $ \to $ Fe(OH)₃(s)+ 3 NaCl(aq).

Moles of Fe(OH)₃ = ${{2.14} \over {107}}$ = 2 $ \times $ 10^$-$2

1 mole of Fe(OH)₃ is obtained from = 1 mole of FeCl₃

$\therefore\,\,\,$ 2 $ \times $ 10^$-$2 moles of Fe(OH)₃ will obtain from

= 0.02 mole of FeCl₃

Molarity of FeCl₃ = ${{No.of\,moles} \over {Volume\,in\,L}}$ = ${{2 \times {{10}^{ - 2}}} \over {0.1}}$ = 0.2 M

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

M₂CO₃ + 2HCl $\buildrel \, \over \longrightarrow $ 2MCl + CO₂ + H₂O

Here weight of M₂CO₃ = 1 gm

Let molar mass of M₂CO₃ = M

$\therefore\,\,\,$ No of moles of M₂CO₃ = ${1 \over M}$

For this reaction,

${{{1 \over M}} \over 1} = {{0.01186} \over 1}$

$ \Rightarrow \,\,\,M$ = ${1 \over {0.01186}}$ = 84.3 g mol^-1

NOTE :

For any reaction this will be always valid.

For this reaction, 2KClO₃ $\buildrel \, \over \longrightarrow $ 2KCl + 3O₂

We can write, ${{{n_{KCl{O_3}}}} \over 2} = {{{n_{KCl}}} \over 2} = {{{n_{{O_2}}}} \over 3}$

This means

We know that 8% sulphur means 8 g of sulphur present in 100 g of an organic compound.

Thus, 32 g of sulphur present in ${{100} \over 8} \times 32g$ = 400 g of organic compound.

Therefore, minimum molecular weight of the compound is 400 g mol^$-$1.

2H₃ As O₄ + 5H₂S  $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{HCl}^{conc}} $   As₂S₅ + 8H₂O

35.5 g of H₃AsO₄ = ${{35.5} \over {142}}$ = 0.25 moles

Let, As₂S₅ produced = n moles.

$\therefore\,\,\,$ ${{0.25} \over 2} = {n \over 1}$

$ \Rightarrow $$\,\,\,$ n = 0.125 mol.

$2$ mole of water softner require $1$ mole of $C{a^{2 + }}\,\,$ ion

So, $1$ mole of water softner require ${1 \over 2}$ mole of $C{a^{2 + }}\,$ ion

Thus, ${1 \over {2 \times 206}} = {1 \over {412}}mol/g\,\,\,$ will be maximum uptake

Given ratio of mass of O₂ and N₂ = 1 : 4
Let mass of O₂ = w
and mass of N₂ = 4w

$\therefore$ Number of moles of O₂ = ${w \over {32}}$
Number of molecules of O₂ = ${w \over {32}}$$ \times $N_A

Number of moles of N₂ = ${4w \over {28}}$
Number of molecules of N₂ = ${4w \over {28}}$$ \times $N_A

$\therefore$ Ratio of molecules of O₂ and N₂
= ${w \over {32}}$$ \times $N_A : ${4w \over {28}}$$ \times $N_A

= 7 : 32

Required reaction,

C_$x$H_$y$ + $\left( {x + {y \over 4}} \right)$O₂ $ \to $$x$CO₂ + ${{y \over 2}}$H₂O

0.72 gm of H₂O = ${{{0.72} \over {18}}}$ mole of H₂O = 0.04 mole of H₂O

In one H₂O molecule 2 hydrogen atoms present.

So in 0.04 mole of H₂O molecules 2$ \times $0.04 = 0.08 moles of H atoms present.

0.72 gm of CO₂ = ${{{3.08} \over {44}}}$ mole of CO₂ = 0.07 mole of CO₂

And in one CO₂ molecule 1 C atom present.

So in 0.07 mole of CO₂ molecules 0.07$\times$1 = 0.07 moles of C atoms present

$\therefore$ C : H = 0.07 : 0.08 = 7 : 8

$\therefore$ Empirical formula of hydrocarbon = C₇H₈

Given metal oxide = M_0.98O
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.

The formula for molarity of mixture of two substance is = ${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$

Here ${M_1} = 0.5$, ${V_1} = 750$, ${M_2} = 2$, ${V_2} = 250$

$\therefore$ Molarity of mixture = ${{0.5 \times 750 + 2 \times 250} \over {750 + 250}}$

= ${{375 + 500} \over {1000}}$

= ${{875} \over {1000}}$

= 0.875

We know molarity (M) = ${{no\,of\,moles\,of\,solute} \over {volume\,of\,solution\,in\,litre}}$

Moles of solute = ${{120} \over {60}}$ = 2

Mass of solution = 1000 + 120 = 1120 gm

Density of solution = 1.15 g/mL

$\therefore$ volume of solution = ${{1120} \over {1.15}}$ = 973.9 mL = ${{973.9} \over {1000}}$ litre = 0.9739 litre

$\therefore$ M = ${2 \over {0.9739}}$ = 2.05

The formula between Mole fraction and Molality is $$m = {{{X_{solute}}} \over {{X_{solvent}}}} \times {{1000} \over {{M_{solvent}}}}$$
$m$ = molality, X_solute = Mole fraction of solute, X_solvent = Mole fraction of solvent, M_solvent = 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

M_solvent = 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

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, M_solute = 98

$\therefore$ 3.6 = ${{10 \times 29 \times d} \over {98}}$

$ \Rightarrow d = 1.22$

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.

In one Mg₃(PO₄)₂ molecule no of oxygen atoms = 8

So if there is M mole of Mg₃(PO₄)₂ molecules then no of moles of oxygen atoms is = M $ \times $ 8

According to question,

M $ \times $ 8 = 0.25

$ \Rightarrow $ M = 0.03125 = 3.125 $\times$ 10^-2

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 (m_solute) = 2.05 $ \times $ 60 = 123

$\therefore$ Mass of solvent (m_solvent) = 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, M_solute = 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

Let Relative Atomic Mass is in short R.A.M then

R.A.M = ${{mass\,of\,1\,atom} \over {{1 \over {12}} \times mass\,of\,1\,atom\,of\,{C^{12}}}}$

$ \Rightarrow $ ${{mass\,of\,1\,atom} \over {mass\,of\,1\,atom\,of\,{C^{12}}}}$ $\times$ 12

Now if we take 1/6 instead of 1/12 at first then the equation will become

$ \Rightarrow $ ${{mass\,of\,1\,atom} \over {mass\,of\,1\,atom\,of\,{C^{12}}}}$ $\times$ 6

as 6 is half of 12 then R.A.M will decrease twice.

Short Cut Method : Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.

Here molarity of first solution is 1.2 and molarity of second solution is 1.5.
So the molarity of the mixture will always be more than 1.2 and less than 1.5. Remember the molarity of mixture can't be either 1.2 or 1.5.

So from the option you an see only 1.344 M can be the right answer.

Normal method :
The formula for molarity of mixture of two substance is = ${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$

Here ${M_1} = 1.5$, ${V_1} = 480$, ${M_2} = 1.2$, ${V_2} = 520$

$\therefore$ Molarity of mixture = ${{1.5 \times 480 + 1.2 \times 520} \over {480 + 520}}$

= ${{720 + 624} \over {1000}}$

= ${{1344} \over {1000}}$

= 1.344

Initially total H₂SO₄ present = 100 mL of 0.1 M = ${{{100} \over {1000}} \times 0.1}$ mole = 0.01 mole

2NaOH + H₂SO₄ $ \to $ Na₂SO₄ + 2H₂O
Let in this reaction H₂SO₄ required n mole

$\therefore$ ${{{{20} \over {1000}} \times 0.5} \over 2}$ = ${n \over 1}$

$ \Rightarrow n$ = 0.005 mole

So now remaining H₂SO₄ = 0.01 - 0.005 = 0.005 mole . Those remaining H₂SO₄ will react with NH₃.

2NH₃ + H₂SO₄ $ \to $ (NH₄)₂SO₄
Let no moles of NH₃ produce through this reaction is = $x$

$\therefore$ ${x \over 2}$ = ${{0.005} \over 1}$

$ \Rightarrow x$ = 0.01

In NH₃ no of N atom is 1 and H atom is 3. So no of moles of N atom in NH₃ = 0.01$ \times $1 = 0.01 mole

This 0.01 mole or 0.01$ \times $14 gm N is produced from 0.3 gm unknown organic compound.

$\therefore$ % of N in unknown compound is

= ${{0.01 \times 14} \over {0.3}} \times 100$

= 46.6

% of N in urea [(NH₄)₂CO] = ${{14 \times 2} \over {60}} \times 100$ = 46.6 %
[ Mol weight of urea = 60]

% of N in benzamide [C₆H₅CONH₂] = ${14 \over {121}} \times 100$ = 11.5 %
[ Mol weight of benzamide [C₆H₅CONH₂] = 121]

% of N in acetamide [CH₃CONH₂] = ${14 \over {59}} \times 100$ = 23.4 %
[ Mol weight of acetamide [CH₃CONH₂] = 59]

% of N in thiourea [NH₂CONH₂] = ${{14 \times 2} \over {76}} \times 100$ = 36.8 %
[ Mol weight of thiourea [NH₂CSNH₂] = 76]

$\therefore$ compound is urea.

Moles (n) = ${{molecules} \over {{N_A}}}$

$ \Rightarrow $ ${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}}$

$ \Rightarrow $ 0.001

Concentration (M) = ${{moles} \over {volume}}$

Given volume of urea is 100 ml = 0.1 litre

[Note : While calculating Concentration volume should always be in litre.]

$ \Rightarrow $ ${{0.001} \over {0.1}}$

$ \Rightarrow $ 0.01 M

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

The reaction is
2BCl₃ + 3H₂ $ \to $ 2B + 6HCL

$\therefore$ 3 moles of hydrogen produces 2 moles of boron

So 2$ \times $10.8 gm = 21.6 gm of boron is produces from 3$ \times $22.4 = 67.2 L hydrogen

Ba(OH)₂ + 2HCl(aq) $ \to $ BaCl₂(aq) + 2H₂O(l)

Assume molarity of Ba(OH)₂ = M

No of moles of Ba(OH)₂ present = 25$\times$M

No of moles of HCl present = 35$\times$0.1

Stoichiometric coefficient of Ba(OH)₂ = 1

Stoichiometric coefficient of HCl = 2

$\therefore$ ${{25 \times M} \over 1}$ = ${{35 \times 0.1} \over 2}$

$ \Rightarrow $ M = 0.07

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 $\times$ N_A

$ \Rightarrow $ 10 N_A

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 $\times$ N_A

$ \Rightarrow $ 5 N_A

Now we can clearly see 10 is twice of 5, so option (A) is correct.

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 fraction of solute =

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.

Element	Weight ratio	No of atoms of element	Simplest Ratio
C	9	9/12 = 3/4	3
H	1	1/1 = 1	4
N	3.5	3.5/14 = 1/4	1

Empherical Formula of the compound (EF) will be C₃H₄N

Now to find molecular formula at first find number of moles of given mass

No. of moles (n) =

Given mass

Molar mass

$ \Rightarrow $ ${{108} \over {54}}$

$ \Rightarrow $ 2

$\therefore$ No. of Moles is 2

Molecular formula (MF) = n (no. of moles) $\times$ Empherical Formula (EF)

$ \Rightarrow $ 2 $\times$ C₃H₄N

$ \Rightarrow $ C₆H₈N₂

$\therefore$ Option (C) is correct