Mathematics `color{red} ✎` PROBLEM SOLVING TECHNIQUES
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Class 11 Chapter 1 - Basic Concepts of Chemistry

`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 from`Fe = 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
 
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