Mathematics color{red} ✎ PROBLEM SOLVING TECHNIQUES
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### Topics covered

• Calculation of No Of Moles
• Balancing of Chemical equation
•Stoichiometry
•Concentration Terms

### PST-1 Calculation of No Of Moles

color{red} ☛ How to Find no. of moles color{red} ("IN")

(A) color{green} ("Given no of Particles")

"no of moles" = ("no of particles")/(N_A (6.022 xx 10^(23))

(B) color{green} ("Given mass of atoms")

"no of moles" = ("mass of atoms in gram")/("molar mass of atom"("grams"// "moles"))

(C) color{green} (" Given mass of compound")

" no of moles " = ("mass of compound in grams")/("molar mass"("grams"//"moles"))

(D) color{green} ("Given volume of any compound at S.T.P")

"no of moles " = ("volume in litre at S.T.P")/(22.4)

Here S.T.P means standard temprature and pressure

Standard temprature = 273K

Standard Pressure = 1 bar => 10^5  pascal (N/m^2)

Earlier standard pressure was taken to be 1 atm . Now standard pressure is 1 bar.

1 atm = 760 mm Hg = 1.01325

= 1.01325 xx 10^5 Pa

1  bar = 1 xx 10^5 Pa

1 atm = 1.01325 bar

1 bar =0.986923 atm.

1 bar equiv 1 atm

(Earlier STP conditions were taken as 1 atm and 00C. But now it is taken as 1 bar and 0^0C. Under these conditions, instead of 22.4L, we have 22.7L.But unless specified consider 1 atm and 0^0C.)

(E) color{green} ("Given mass of ionic compound")

"no of moles" = ("mass of ionic compound in grams")/("Formula Mass")
Q 3115391260

How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second -

(A)

19.098 xx 10^19 years

(B)

19.098 years

(C)

19.098 xx 10^9 years

(D)

None of these.

Solution:

because 10^6 rupees are spent in 1 sec.

:. 6.022 xx 10^23 rupees are spent in = (1 xx 6.022 xx 10^23)/10^6  sec

= (1xx 6.023 xx 10^23)/( 10^6 xx 60 xx 60 xx 24 xx 365 ) years

= 19.098 xx 10^9 year
Correct Answer is => (C) 19.098 xx 10^9 years
Q 3156212174

A piece of Cu contains 6.022 xx 10^24 atoms. How many mole of Cu atoms does it contain?

Solution:

No. of mole = (6.022 xx 10^24)/(N_A) =(6.022 xx 10^24)/(6.022 xx 10^23) = 10 mole
Q 3155391264

Calculate the number of Cl^- and Ca^+2 ions in 222 g anhydrous CaCl_2.

(A)

2N_A ions of Ca^(+2) 4 Na ions of  Cl^-

(B)

2N_A ions of Cl^- & 4Na ions of Ca^(+2)

(C)

1 N_A ions of Ca^(+2) & 1 Na ions of Cl^-

(D)

None of these.

Solution:

because mol. wt. of CaCl_2 = 111 g

because 111 g CaCI_2, has = N_A ions of Ca^(+2)

:. 222g of CaCl_2 has (N_A xx 222)/111 = 2 N_A ions of Ca ^(2+) .

Also because 111 g CaCl_2 has = 2N_A ions of Cl^-

:. 222 g Ca Cl_2 has  = (2 N_A xx 222)/111 ions of Cl^- = 4N_A ions of Cl^-
Correct Answer is => (A) 2N_A ions of Ca^(+2) 4 Na ions of  Cl^-
Q 3105391268

What is the molecular mass of a substance, each molecule of which contains 9 carbon atoms, 13 hydrogen atoms and 2.33 xx 10^-23 g of other component?

Solution:

The molecule has C, H and other component.
.
Mass of 9 C atoms = 12 xx 9 = 108 amu
Mass of 13 H atoms = 13 xx 1 = 13 amu

Mass of other component = (2.33 xx 10^-23)/(1.66 xx 10^-24) = 14.04 amu

:. Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu

:. Mol. mass of substance = 135.04
Q 3115491360

The density of O_2 at 0^oC  and 1 atm is 1.429 g/litre. The molar volume of gas is-

(A)

22.4 lit.

(B)

11.2 lit

(C)

33.6 lit

(D)

5.6 lit.

Solution:

because 1.429 gm of O_2 gas occupies volume = 1 litre.

:. 32 gm of 0 2 gas occupies  = 32/1.429 = 22.4 litre/mol
Correct Answer is => (A) 22.4 lit.
Q 3186312277

A sample of He gas occupies 5.6 litre volume at 1 atm and 273 K . How many mole of He are present in the sample?

Solution:

No. of mole = (5.6)/(22.4) =0.25

### PST-2 Balancing of Chemical equation

1. First balance those elements which have subscripts

2. O & H are balanced in last between O & H use proper that which is less no of reactants and products.

3. If coefficient in negative fraction multiply all by suitable factor to remove fraction.
Examples

Ba_3N_2 + H_2O -> Ba (OH)_2+ NH_3

CaCl_2+Na_3PO_4->Ca_3(PO_4)_2+NaCl

FeS+O_2->Fe_2O_3+SO_2

PCl_5+H_2O->H_3PO_4+HCl

As+NaOH->Na_3AsO_3+H_2

Hg(OH)_2 + H_3PO_4 ->Hg_3(PO_4)_2 +H_2O

HClO_4+P_4O_10->H_3PO_4+Cl_2O_7

CO+H_2->C_8H_18 +H_2O

KClO_3+P_4->P_4O_10 +KCl

SnO_2+H_2->Sn+H_2O

KOH+H_3PO_4->K_3PO_4+H_2O

KNO_3+H_2CO_3->K_2CO_3+HNO_3

Na_3PO_4+HCl->NaCl+H_3PO_4

TiCl_4+H_2O-> TiO_2+HCl

C_2H_6O +O_2->CO_2+H_2O

Fe+HC_2H_3O_2->Fe(C_2H_3O_2)_3+H_2

NH_3 +O_2->NO +H_2O

B_2Br_6 +HNO_3->B(NO_3)_3 +HBr

NH_4 OH +KAI(SO_4)_2 * 12 H_2O ->Al(OH)_3 + (NH_4)_2 SO_4 +KOH + H_2O

### PST-3 Stoichiometry

tt (( H_2 (g) + , I_2 (g) -> , 2 HI(g)), ( 1 " molecule" , 1 " molecule" , 2 " molecule"),( 1 " mole", 1 " mole" , 2 " mole") , ( 1 " volume", 1 " volume", 2 " volume" (T & P " Constant")), ( 1 \ Pressure , 1 \ Pressure , 2 \ Pressure(T & V " constant")))
Q 3156145074

What volume of oxygen at STP is required to affect complete combustion of 200 cm^3 of acetylene and what would be the volume of carbon dioxide formed ?

Solution:

The chemical equation representing the combustion of acetylene is

underset(2 Vol.) (2C_2H_2) +underset(5 Vol)(5O_2) → underset(4 Vol.)(4CO_2) +2H_2O

Step 1 : To calculate the volume of O_2 at STP required to affect complete combustion of 200 cm^3 of acetylene.

Applying Gay Lussac's Law of gaseous volumes 2Vol.  of C_2H_2 require O_2 for complete combustion = 5 Vol.

therefore 200 cm^3 of C_2H_2 will require O_2 for complete combustion  = 5/2 xx 200 = 500cm^3  at STP

Thus the volume of O_2 required = 500 cm^3  at STP

Step 2 : To calculate the volume of CO_2 produced at STP

Applying Gay Lussac's Law of gaseous volumes

2Vol.  of C_2H_5 produced CO_2 = 4Vol.

therefore 200 cm^3 of C_2H_2 at STP will produce CO_2

 = 4/2xx200 = 400 cm^3  at STP

Thus the volume of CO_2 produced

= 400 cm^3 at STP
Q 3116034879

1.0g of a mixture of carbonates of calcium and magnesium gave 240 cm^3 of CO_2 at STP. Calculate the percentange composition of the mixture.

Solution:

Mass of mixture of carbonates of Ca and Mg taken = 1.0 g

Let the mass of CaCO_3 = xg

therefore Mass of MgCO_3 = (1-x)g

The chemical equations involved are :

underset(40+12+3xx16 = 100g)(CaCO_3) → CaO +underset(22400 cm^3 text(at) ATP)(CO_2) ...........(i)

underset(24+12+3xx16= 84g)(MgCO_3) → MgO + underset(22400 cm^3 text(at) STP) (CO_2) ............(ii)

Step 1 : To calculate the volume of CO_2 evolved at STP from xg of CaCO_3

100g of CaCO_3 will evolved of CO_2 at STP = 22400 cm^3

therefore xg of CaCO_3 will evolved CO_2 at STP  = 22400/100xx x cm^3

Step 2 : To calculate the volume of CO_2 evolved at STP from (1-x)g  of MgCO_3.

84g of  MgCO_3 evolve CO_2 at STP = 22400 cm^3

therefore (1-x)g of MgCO_3 will be evolve CO_2 at STP.

= 22400/84 xx (1-x) cm^3 = 800/3 ( 1-x) cm^3

Step 3 : To calculate the volume of x

therefore Total volume of CO_2 evolve at STP

224x+800/3 ( 1-x) cm^3

But total volume of CO_2 evolved at STP

 = 240 cm^3 (given)

therefore 224x+800/3 (1-x) = 240.

or 672x+800-800x = 720

therefore 128x = 80

x = 5/8

Step 4 : To calculate the percentage composition of the mixture.

therefore Percentage of CaCO_3 = 5/(8xx1) xx 100

 = 62.5

therefore Percentage of MgCO_3 = 100-62.5 = 37.5
Q 3186034877

Current market prices of Al , Zn and Fe scraps per kg are Rs. 20, Rs. 16 and Rs. 3 respectively. If H_2 is to be prepared by the reaction of one of these metals with H_2SO_4 , which would be the cheapest metal to use ? Which would be most expensive? .

Solution:

The various chemical reactions involved are given below.

(i) underset(2xx27 = 54g)(Al) +3H_2SO_4 → Al_2(SO_4)_3 +underset(3xx2 = 6g)(3H_2)

(ii) underset(65 g)(Zn) +H_2SO_4 →ZnSO_4 +underset(2g)(H_2)

(iii) underset(56g)(Fe) +H_2SO_4 → FeSO_4 +underset(2g)(H_2)

Let us suppose that the amount of hydrogen to be prepared = 100g

Step 1 : To calculate the cost of preparation of 100g of H_2 from Al

6g of H_2 is prepared from Al =54g
therefore  100g of H_2 will be obtained from Al = 54/6 xx 100 = 900g

Cost of 1000 g of Al = Rs. 20

therefore  cost of 900g of Al = 20/1000xx900 = Rs. 18

Step 2 : To calculate the cost of preparation of 100g of H_2 from Zn

2g of H_2 is produced from Zn = 65 g

therefore 100g of H_2 will be obtained from Zn = 65/2xx100= 3250g

Cost of 1000g of Zn = Rs. 16

therefore Cost of 3250 g of Zn = 16/1000 xx3250 = Rs. 52

Step 3 : To calculate the cost of preparation of 100g of H_2 from Fe

2g of H_2 is produced from Fe = 56 g

therefore 100g  of H_2 will be obtained fromFe = 56/2xx100 = 2800g

Cost of 1000g of Fe = Rs. 3

therefore cost of 2800g of Fe = 3/1000 xx 2800 = Rs. 8.40

Thus Fe is the cheapest and Zn is the most expensive metal to use from the preparation of H_2
Q 3106834778

Calculate the mass of iron which will be converted in to its oxide (Fe_3O_4) by the action of 18 g of steam on it.

Solution:

The chemical equation representing the reaction is

tt ((3Fe , + , 4H_2O , → , Fe_3O_4 , + , 4H_2) , (3xx56 , , 4xx18 , , , ,), (=168g , , = 72 g , , , ,) )

Thus , 72g of steam react with 168g of iron .

therefore  18 g of steam will react with 168/72 xx 18 = 42 g of iron.

therefore  Mass of iron required = 42 g

### PST-4 Concentration Terms

color{green}☛ M= (md)/(1+ mM_2//1000)

color{green}☛ d= M (1/m + M_2 /1000)

color{green} ☛ m= (1000 xx x_2)/(x_1 M_1)

color{green} ☛ M= (1000 d x_2)/(x_1M_1 +x_1M_2)

M= Molarity
d= Density of solution
x_2 = Mole fraction of solute
M_1= Molecular mass of solvent

m= molarity
M_2= Molecular mass of solute
x_1= Mole fraction of solvent
d= Density of solution.
Q 3135191962

When measured at the same temperature and pressure, hydrogen reacts with oxygen to form water in the volume ratio 2 : 1. Calculate the volume of O_2 gas measured at 137°C and 760 mm pressure that will combine with 100 ml of H_2 at 0°C and 200 mm pressure.

Solution:

Reaction: underset (2 vol ) (2H_2) + underset (1 Vol) (O_2) -> 2H_2O

2 volume of H_2 required O_2 = 1 Vol

:.  100 mL of H_2 required O_2 = 100/2 =50 ml at 0°C and 200 mm

:. P_1  = 200 mm , V_1 =50 mL , T_1 = 0 +273 K , P_2 = 760 mm

V_2 = ? , T_2 = 137 + 273 = 410 K

But (P_1V_1)/T_1 = (P_2 V_2)/T_2 gas equation

Hence , V_2 = (P_1V_1T_2)/(P_2T_1)

:. V_2 = (200mm xx 50 mL xx 410 K)/( 760mm xx 273 K ) = 19.76  ml
Q 3146212173

Derive the relation between molality (m) and mole fraction of solute, chi_2

Solution:

Molality, m means, m mole of solute in 1000 g of solvent which is equal to 1000//M_1  mol
where M_1 =  molar mass of the solvent.

:.  Mole fraction , chi_2 = text(moles of solute)/text( Moles of solute+ Moles of solvent)

= m/ ( m + 1000/M_1) = (mM_1 )/ ( mM_1 + 1000)

Hence  m = (1000 x chi_2 )/(1 -X_2)
Q 3116312270

The molality and molarity of a solution of H_2SO_4 are 94.13 and 11.12 respectively. Calculate the density of the solution.

Solution:

d = M ( 1/m + (mol.wt.)/1000 )

= 11.12 ( 1/94.13 + 98/1000)

= 1.2079 g/ml