Some Basic Concepts Of Chemistry - Class 11 Engineering Chemistry - Extra Questions

Define a) Atomic mass unit
            b)Mole

Atomic mass unit:
A unit of mass used to express atomic and molecular weights, equal to one-twelfth of the mass of one atom of carbon $$12$$. It is also known as Dalton

Mole:-
Standard scientific unit for meaning large quantities of very small entities such as atoms molecular or other specified particles.
The mole is the unit of measurement for the amount of substance in the International System of Unit $$(SI)$$. It is defined as the amount of a chemical substance that contains exactly $$6.02214076 \times 10^{23}$$ (Avogadro's constant) consecutive particles eg atoms. molecules, ions, or electrons.


Define mole.

$$Explanation:$$

 Mole can be defined as:

The amount of substance that contains the same number of chemical species (atoms, molecules, ions &, etc.) as the number of atoms present in $$12$$g of the carbon-$$12$$ isotope is considered as a 1-mole substance.

or

The amount of substance that contains  $$6.02 \times 10^{23}$$ atoms/ions/molecules is called a 1-mole substance.
or


The number of gas molecules present in $$22.4$$ liter of volume at $$STP$$ is called 1 mole. 



How many atoms of hydrogen are there in 36 g of $$NH4^+$$?

Molar mass of $$NH_4^+$$ $$= 14+4= 18 \ g\ mole^{-1}$$

No. of moles of  $$NH_4^+$$= $$\dfrac {given \ mass}{molar \ mass}= $$ $$\dfrac{36}{18} = 2 \ moles$$

Now total moles of hydrogen atoms $$ = 4 \times 2 \ moles=$$ $$8 \times 6.022 \times 10^{23}= 48.176 \times 10^{23} \ or \ 4.82 \times 10^
{24}.$$



If $$100$$ grams of calcium carbonate (whether in the form of marble or chalk ) are decomposed completely, then $$56$$ grams of calcium oxide and $$44$$ grams of carbon dioxide are obtained. Which law of chemical combination is illustrated by this statement?

This reaction illustrates the law of conservation of mass i.e.mass before the decomposition is the same as mass after the decomposition $$(i.e. 56+44 =100g)$$


Explain Gay Lussac's Law of combining volume with example .


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If $$100$$ grams of pure water taken from different sources is decomposed by passing electricity. $$11$$ grams of hydrogen and $$89$$ grams of oxygen are obtained. Which chemical law is illustrated by this statement?

This reaction illustrates the law of conservation of mass i.e. mass before the decomposition is the same as mass after the decomposition $$(i.e. 89+11 =100g)$$
The reaction is as follow:
$$2H_2O \rightarrow 2H_2 + O_2$$


$$1\ g$$ of $$N_2$$ and $$1\ g$$ of $$O_2$$ are put in a two-litre flask at $$27^oC$$. Calculate partial pressure of each gas, the total pressure and the composition of the mixture in mole percentage.


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What is meant by a mole of a substance?

The mole (mol) is the unit of measurement for the amount of substance in the International System of Units (SI). A mole of a substance or a mole of particles is defined as exactly $$6.022×10^{23} $$ particles, which may be atoms, molecules, ions, or electrons.
A mole of carbon atoms  $$= 6.022×10^{23}$$ carbon atoms. 


Give short answers:
What is alchemy?

The word "Alchemy" has its origin in a Greek word 'Khemeia' means 'art of transmuting metals'.It was partly based on experimentations and partly on spiritual discipline. 


Experiment and _____________ are the two important basics of chemistry.

observation


Define :
Relative atomic mass

Relative atomic mass is the mass of an atom of an element as a multiple of the standard atomic mass unit.


Sodium chloride reacts with silver nitrate to produce silver chloride and sodium nitrate. State the law that satisfies this equation.

$$NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$$

This equation states law of conservation of mass where mass is neither created nor destroyed.


Which isotope is used as a reference on the atomic scale? What is one amu or one 'u' ?

  • $$Carbon - 12$$ isotope is the most abundant isotope of carbon and has been chosen as standard or reference on the atomic scale
  • One atomic mass unit is defined as one-twelfth of the mass of one $$carbon - 12$$ atom because $$C-12$$ is chosen as the standard atom, and this is because it has an equal number of protons and neutrons (6) and constitutes most of the matter. Also the most abundant isotope of carbon is $$Carbon - 12.$$


Number of moles present in 28g of nitrogen atoms are

$$\textbf{Given data:}$$

Given mass of nitrogen atom = 28 g

$$\textbf{Step: Calcuate the number of moles present in 28 g of nitrogen atoms.}$$


$$Number\ of\ moles\ of\ nitrogen = \dfrac{Given\ mass\ of\ nitrogen}{Atomic\ mass\ of\ nitrogen}$$


$$Number\ of\ moles\ of\ nitrogen = \dfrac{28}{14}$$


$$Number\ of\ moles\ of\ nitrogen = 2 \ mol$$


$$\textbf{Conclusion:}$$

Therefore, the number of moles present in 28 g of nitrogem atoms are 2 mole.



What is the SI definition of mole?

The mole is the SI unit of amount of substance. One mole contains exactly $$6.02214076\times 10^{23}$$


In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

The molar mass of $$\displaystyle BaSO_4 $$ is 233 g/mol.
Thus, 233 g of $$\displaystyle BaSO_4 $$ corresponds to 32 g of sulphur.
0.668 g of $$\displaystyle BaSO_4 $$ will contain $$\displaystyle \frac {32 \times 0.668}{233} = 0.0917g $$ of sulphur.
The percentage of  sulphur in the given compound is $$\displaystyle \frac {0.0917 \times 100}{0.468} = 19.6$$ %.


At STP, number of molecules in $$5.6 l$$ of sulphur trioxide gas is _________.

At STP one mole of gas occupies $$22.4\ L$$.
So, no of molecules in $$5.6l$$ of $$SO_3$$ are $$\dfrac{5.6}{22.4} \times Na=1.5\times10^{23}$$


Calculate the number of molecules of ammonia present in $$5.6\ dm^3$$ of its volume.

$$V = 5.6\ dm^3 = 5.6\ L$$

Then ,
We know that,

$$n = V / 22.4 $$
$$n = 5.6 / 22.4$$
$$n = 56 / 224$$
$$n = 0.25$$ mole.
And for the no of molecules,

We have a concept,

No. of molecules $$= n$$ $$\times$$ Avogadro number
No. of molecule $$=0.25\times 6.022\times 10^{23}$$
No. of molecules $$=1.5 \times 10^{23}$$



Calculate the mass of the following
i) 0.5 moles of $$N_2$$ gas
ii) 0.5 moles of N atom 

$$(i)$$ Mass of $$N_2=$$ Moles of $$N_2\times $$ Molecular weight of $$N_2=0.5\times 28=14$$ $$g$$
$$(ii)$$ Mass of $$N-$$ atom $$=$$ Moles of $$N-$$ atom $$\times $$ Molecular weight of $$N-$$ atom $$=0.5\times 14=7$$ $$g$$


Calculate the number of moles for the following:
i) 52 g of He
ii) number of He atom 

Given mass $$=52 g$$
Molecular mass of He $$=4 g$$
$$(i)$$ No. of moles $$=\cfrac{Given\quad mass}{Molecular\quad mass}=\cfrac{52}{4}=13$$ moles

$$(ii)$$ No. of atoms $$=$$ No. of moles of $$He$$ atom $$\times$$ Avogadro's no. $$=13\times 6.022\times 10^{23}=783\times 10^{24}$$ atoms



How is $$1$$ mole NaOH different from $$1$$ molar NaOH.

1 Mole $$NaOH$$ means 40 grams of $$NaOH$$.
1 Molar $$NaOH$$ means 40 grams of $$NaOH$$ is dissolved in 1000 ml of water.


Convert the following into moles: 
i) 12g of $${ O }_{ 2 }$$ gas
ii) 20g of water
iii) 22g of $${ CO }_{ 2 }$$
iv) 52g of helium

Formula of mole: 

$$mole(n) = \dfrac{Given \ mass}{Molar \ mass}$$
 
$$12g$$ of $$O_{2}\Rightarrow \dfrac{12}{32}=\dfrac{3}{8}mol$$ 

$$20g $$ of $$H_{2}O\Rightarrow \dfrac{20}{18}=\dfrac{10}{9}mol$$

$$22g$$ of $$CO_{2}\Rightarrow \dfrac{22}{44}=0.5mol$$

$$52g$$ of $$He\Rightarrow \dfrac{52}{4}=13mol$$


Calculate the percentage composition of each element in Potassium chlorate, $$KClO_3.$$

Relative molecular mass of $$KClO_3$$
$$= 39.09 + 35.5 + (3 \times 16)$$
$$= 122.59 \,g$$
$$122.59 \,g \,KClO_3$$ contains $$39.09 \,g \,K$$
Hence, $$100 \,g \,KClO_3$$ contains
$$= \dfrac{100 \times 39.09}{122.59} = 31.9 \,g$$
$$122.59 \,g \,KClO_3$$ contains $$35.5 \,g \,Cl$$
Hence, $$100 \,g \,KClO_3$$ contains
$$= \dfrac{100 \times 35.5}{122.59} = 28.9 \,g$$
$$122.59 \,g \,KClO_3$$ contains $$48 \,g \,O$$
Hence, $$100 \,g \,KClO_3$$ contains
$$= \dfrac{100 \times 48}{122.59} = 39.1 \,g$$
The percentage of $$K, Cl$$ and $$O$$ in $$KClO_3$$ are $$31.9\%, 28.9\%$$ and $$39.1 \%$$ respectively.


Urea is a very important nitrogenous fertilizer. Its formula is $$CON, H_4.$$ Calculate the percentage of carbon in urea. $$(C = 12, O = 16, N = 14 \,and \,H = 1)$$

ElementNo of atomsAtomic massTotal
$$H$$$$2$$$$14$$$$28$$
$$C$$$$1$$$$12$$$$12$$
$$H$$$$4$$$$1$$$$4$$
$$O$$$$1$$$$16$$$$16$$
$$[12 + 16 + 28 + 4 = 60]$$
Hence, relative molecular mass of urea $$= 60$$
Percentage of carbon = weight of carbon / Total weight of urea $$\times 100$$
$$= 12 / 60 \times 100$$
$$= 20$$ or $$20\%$$


Fill in the blank.
Isotopes are the atoms of an element having the same  _________ number but a different __________ number.

Isotopes are the atoms of an element having the same atomic number but a different mass number as they contain the same number of electrons and protons but a different number of neutrons.


$$ 0 \cdot 5 \mathrm{g} $$ mixture of $$ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} $$ and $$ \mathrm{KMnO}_{4} $$ wastreated with excess KI in acidic medium. Iodine liberated required $$ 100 \mathrm{cm}^{3} $$ of $$ 0 \cdot 15 \mathrm{N} $$ sodium thiosulphate solution for titration. Find the percent amount of each in the mixture (At. wts. K $$ =39, \mathrm{Cr}=52, \mathrm{Mn}=55, \mathrm{Na}=23 \mathrm{S}=32) $$

The reactions involved are:
$$ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+7 \mathrm{H}_{2} \mathrm{SO}_{4}+6 \mathrm{Kl} \longrightarrow \quad 4 \stackrel{294 \mathrm{g}}{\mathrm{K}_{2} \mathrm{SO}_{4}}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{I}_{2} \quad \dots(i)$$
$$\begin{array}{l}3 \times 254 \\=762 \mathrm{g}\end{array}$$
$$ 2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{KI} \rightarrow $$
$$2 \times 158$$
$$ =316 \mathrm{g} $$
$$ \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{I}_{2} \quad \dots(ii)$$
$$\begin{array}{l}5 \times 254 \\=1270 \mathrm{g}\end{array}$$
$$ 2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI} $$
Suppose weight of $$ \mathbf{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} $$ in the mixture $$ =x g $$
Then weight of $$ \mathrm{KMnO}_{4} $$ in the mixture
$$=(0 \cdot 5-x) g$$
Iodine produced from $$ x g K_{2} C r_{2} O_{7}=\dfrac{762}{294} \times x g $$
Iodine produced from $$ (0 \cdot 5-x) g \mathrm{KM} \mathrm{n} \mathrm{O}_{4} $$
$$=\dfrac{1270}{316} \times(0 \cdot 5-x) g$$
Total iodine produced
$$=\dfrac{762 x}{294}+\dfrac{1270(0 \cdot 5-x)}{316} g$$
$$ 100 \mathrm{cc} $$ of $$ 0 \cdot 15 \mathrm{N} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} $$
$$ =100 \mathrm{cc} $$ of $$ 0 \cdot 15 \mathrm{N} \mathrm{I}_{2} $$ solution
$$=\dfrac{127 \times 0 \cdot 15}{1000} \times 100 g=1 \cdot 905 g$$
$$ \operatorname{Hence} \dfrac{762 x}{294}+\dfrac{1270(0 \cdot 5-x)}{316}=1 \cdot 905 $$
which on solving gives $$ x=0.073 \mathrm{g} $$ 
$$ \therefore \% $$ age of $$ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} $$ in the mixture
$$ =\dfrac{0 \cdot 073}{0 \cdot 5} \times 100=14 \cdot 6 \% $$
and $$ \% $$ age of $$ \mathrm{KMnO}_{4} $$ in the mixture $$ 100-14 \cdot 6=85 \cdot 4 \% $$


18.4g mixture of $$CaCO_3$$ and $$MgCO_3$$, on strongly heating gives 8.8g of $$CO_2$$ gas. Calculate mole percentage of $$MgCO_3$$ in the mixture.

Let $$CaCO_3\, be\, x g$$.
Then, $$MgCO_3\, is\, (18.4 - x)g$$.
$$CaCO_3\rightarrow CaO+CO_2$$
1 mol            1 mol     1 mol
100g             56g        44g
xg                                $$\frac{44x}{100}$$g

$$MgCO_3\rightarrow MgO+CO_2$$
1 mol             1 mol     1 mol
84g                40g       44g
(18.4-x)g                 $$\frac{44(18.4-x)}{84}g$$

Total amount of $$CO_2=\displaystyle \frac{44x}{100}\, +\, \frac {44(18.4\, -\, x)}{84}$$ = 8.8 (given).

This gives x = 10g.

$$\therefore\, CaCO_3\, =\, 10 g = 0.1 mol$$.

$$MgCO_3\, =\, 18.4\, -\, 10\, =\, 8.4g\, =\, 0.1 mol$$.

Total moles of $$CaCO_3$$ and $$MgCO_3\, =\, 0.2\, mol$$.

$$\therefore\, CaCO_3\, =\, 50\,mol\%$$.

$$MgCO_3\, =\, 50\,mol\%$$.


If the water molecules in 1.0 g of water were distributes uniformly over the surface of earth, If there are $$6.5\times10^{-x}$$such molecules would there be in $$1.0\, cm^2$$ of earth's surface , then what is the value of x?

Number of water molecules in one gram of water is 

$$N=\dfrac { 1 }{ 18 } { N }_{ A }=\dfrac { 6.022X{ 10 }^{ 23 } }{ 18 } =3.35X{ 10 }^{ 22 }$$

Distributing them evenly over the surface of earth (R=6400km) gives areal density

$$\rho =\dfrac { N }{ A } =\dfrac { N }{ 4\pi { R }^{ 2 } } =\dfrac { 3.35X{ 10 }^{ 22 } }{ 4(3.142)({ 640000000 }^{ 2 }) } =6500$$

Thus, molecules in unit area are,

$$dN=\rho (1{ cm }^{ 2 })=6.5X{ 10 }^{ 3 }$$

Answer is $$6.5X{ 10 }^{ 3 }$$


The melting points of four solids A, B, C and D are $$78{\,^o}C,\,213{\,^o}C,100{\,^o}C$$ and $$154{\,^o}C.$$ Arrange them in increasing order of their interparticle forces of attraction.

Here $$A = 78{\,^o}C,\,\,\,B =213{\,^o}C,\,\,\,C = 100{\,^o}C,\,\,D = {\,^o}C,154{\,^o}C$$ have melting points.

As stronger the faces of attraction between the particles more will be melting point and vice-versa.

Hence,

Order of their inter-particle forces of attraction is $$B > D > C > A$$ maximum melting point, so maximum inter particles forces of attraction.



Two small iron particles, each of mass $$280 mg $$ , are placed at a distance $$10 cm $$ apart. If $$0.01 \% $$ of the electrons of one particle are transferred to the other, find the electric force between them. Atomic weight of iron is $$56 g mol^{-1} $$ and there are $$26$$ electrons in each atom of iron.

The Avogadro’s number is $$6.022 \times {10^{23}}$$

So in $${\rm{280 mg }}$$the number of atoms are $$\dfrac{{6.022 \times {{10}^{23}} \times 280 \times {{10}^{ - 3}}}}{{56}} = 3 \times {10^{27}}$$

The total number of electron in $$3 \times {10^{27}}$$atoms will be $${\rm{3 \times 1}}{{\rm{0}}^{{\rm{27}}}}{\rm{ \times 26 e}}$$

The charge transferred is $$\dfrac{{\rm{1}}}{{{\rm{100}}}}{\rm{ \times 78 \times 1}}{{\rm{0}}^{{\rm{27}}}}{\rm{e}}$$ are transferred to the other atom placed at distance of $$10cm$$ the force will be

$$F = k{\dfrac{{{Q_1} \times {Q_2}}}{{{r^2}}}^{}}$$

$$F = \dfrac{{ - 1.40 \times {{10}^{26}}}}{{0.01}}$$

$$F =  - 1.40 \times {10^{28}}$$

 



On passing $$25 \ mL$$ of a gaseous mixture of $$N_2$$ and $$NO$$ over heated copper, $$20 \ mL$$ of the gas remained. Calculate the percentage of each in the mixture.

$$V(N_2 + NO) = 25 mL $$

Let volume of $$N_2 = x mL$$

volume of $$NO = (25 - x) mL$$

$$Cu  +  \underset{(25-x)}{NO}  \rightarrow \ CuO  +  \dfrac{1}{2} N_2 (g) $$ 

($$N_2$$ is highly uncreative due to very strong triple bond  it only react with Mg, Li etc)

Volume of $$N_2$$ produced from $$NO$$ gas = $$\dfrac{(25 -x)}{2}$$

Volume of $$N_2$$ present from before $$= x \ mL$$

$$\therefore$$ A/q, $$x + \dfrac{(25 - x)}{2} = 20 $$

$$ 2x + 25 -x = 40$$

$$ X = 40 -25 = 15 mL$$

$$\therefore$$ Volume of $$N_2 = x = 15 mL$$

$$\therefore \% \ of \ N_2 = \dfrac{15}{25} \times 100  = 60 \%$$ 
 
$$\therefore \% \ of \ NO = \dfrac{10}{25} \times 100  = 40 \%$$ 



Excess of KI and dil  $$ H_2SO_4 $$ were mixed in 50 mL $$ H_2O_2 $$ The liberated 12 required 20 mL of 0.1 N $$ Na_2S_2O_3 $$ Find out the strength of $$ H_2O_2 $$ in g/litre.

  • $$I_2​+2Na_2​S_2​O_3​\rightarrow {}Na_2​S_4​O_6​+2NaI$$

20ml of 0.1M Na2​S2​O3​ will react with $$\frac{(20)(0.1)}{​(1000)(2)}$$ moles of $$I_2$$​ which is $$0.001\; moles \;or\; 0.002\; equivalents$$.

Thus there are 0.002 equivalents in $$50ml$$ of $$H_2​O_2$$​.

Hence, its strength is $$0.002 \;X\; 17/0.05=0.68$$  g/L {Answer}


$$40 \ mL$$ of a mixture of hydrogen, $$CH_4$$ and $$N_2$$ was exploded with $$10 \ mL$$ of oxygen. On cooling, the gases occupied $$36.5 \ mL$$. After treatment with $$KOH$$, the volume reduced by $$3 \ mL$$ and again on treatment with alkaline pyrogallol, the volume further decreased by $$1.5 \ mL$$. Determine the composition of the mixture.

$$ V(H_2 + CH_4 + N_2) = 40 mL$$

$$ V_{O_2} = 10 mL$$

$$ H_2 + \dfrac{1}{2} O_2  \rightarrow  \ H_2O(l)$$

$$ CH_4 + \left(1+\dfrac{4}{4}\right) O_2  \rightarrow \ CO_2 + 2H_2O(l)$$

$$ N_2 \rightarrow $$  No reaction 

Let volume of $$H_2 = x mL$$

Volume of $$CH_4 = y \ mL$$

volume of $$N_2 = \{40 - (x + y)\}$$

On cooling the gas remained 
$$= CO_2 (g) $$ produced $$ +  N_2(g)  +  O_2$$ (remained)

A/q, $$ y + \{40 - (x + y)\} + 10 - \left(\dfrac{x}{2} + 2y\right) = 36.5$$

(with x mL of $$H_2, \dfrac{x}{2}$$  mL of $$O_2$$ reacts & with y mL of $$CH_4 \ 2y \ mL$$ of $$O_2$$ reacts)

$$ 40 - x + 10 - \dfrac{x}{2} - 2y = 36.5$$

$$50 - \dfrac{3x}{2} - 2y = 36.5 $$

$$ 50 - 36.5 = \dfrac{3x}{2} + 2y$$

$$\therefore \dfrac{3x}{2} + 2y = 13.5$$

$$3x + 4y = 27$$ ---------------(1)

After passing over KOH, $$CO_2$$ will be absorbed
$$\therefore y = 3 \ ml$$ -------------(2)

After passing over pyrogallol, $$O_2$$ will be absorbed.

$$\therefore \ 10 - \left(\dfrac{x}{2} + 2y \right) = \dfrac{3}{2}$$

$$ 10 - \dfrac{3}{2} = \dfrac{x}{2} + 2y$$

$$ 17 = x + 4y$$ ------------------(3)

 from (2) & (3) $$x = 5$$

$$\therefore$$ Volume of $$H_2  = 5 mL$$

Volume of $$NH_4 = 3 mL$$

Volume of $$N_2 = 32 mL$$

$$\therefore \% \ of \ CH_4 = \dfrac{3}{40} \times  100 = 7.5\%$$

$$\therefore \% \ of \ H_2 = \dfrac{5}{40} \times  100 =12.5\%$$

$$\therefore \% \ of \ N_2 = \dfrac{32}{40} \times  100 = 80\%$$


(i) A solution of a mixture of $$KCl$$ and $$KOH$$ was neutralised with $$120mL$$ of $$0.12N$$ $$HCl$$. Calculate the amount of $$KOH$$ in the mixture.
(ii) After titration, the resultant solution was made acidic with $$H{NO}_{3}$$. Then excess of $$Ag{NO}_{3}$$ solution was added to precipitate the $$AgCl$$ which weighed $$3.7g$$ after drying. Calculate percentage of $$KOH$$ in the original mixture.

$$KCl+KOH+HCl\rightarrow KCl+{H}_{2}O$$
                           $$120ml$$
                           $$0.12N$$

meq of $$KOH=$$ meq. of $$HCl$$ (Only $$KOH$$ will react with $$HCl$$)

$$\cfrac{wt}{eq.wt\quad of\quad KOH}\times 1000=120\times 0.12$$

$$\cfrac{wt}{56/1}\times 1000=120\times 0.12$$

$$wt=\cfrac{120\times 0.12\times 56}{1000}=0.8064g$$

(ii) $$KCl+Ag{NO}_{3}\xrightarrow [  ]{ H{ NO }_{ 3 } } AgCl\downarrow +Ag{NO}_{3}$$

total
$$\cfrac { wt.\quad of\quad { KCl }_{ total } }{ 74.5/1 } =\cfrac{3.7}{143.5}$$

Wt of $${KCl}_{total}=\cfrac{3.7\times 74.5}{143.5}=1.921g$$

Also wt of $$KCl$$ produced from $$KOH=\cfrac{0.8064}{56}\times 74.5=1.0728g$$

wt of $$KCl$$ originally present $$=1.921-1.0728=0.8482g$$

tota wt. of initially $$=0.8482+0.8064=1.6546g$$

% of $$KOH=\cfrac{wt.KOH}{totalwt.}\times 100=\cfrac{0.8064}{1.6546}\times 100=48.74$$%


$$38 \ mL$$ of a mixture of $$CO$$ and $$H_2$$ was exploded with $$31 \ mL$$ of $$O_2$$. The volume after the explosion was $$29 \ mL$$ which reduced to $$12 \ mL$$ when shaken with $$KOH$$. Find the percentage of $$CO$$ and $$H_2$$ in the mixture.


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Match the column $$I$$ with column $$II$$:


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$$ 2.5\,g $$ of a mixture of $$ BaO $$ and $$ CaO $$ when treated with an excessof $$ H_2SO_4 $$ , produced $$ 4.713\,g $$ of the mixed sulphates . Find the percentage of $$ BaO $$ present in the mixture .

$$ W ( BaO + CaO ) = 2.5\,g $$ 

Let $$ x\,g $$ of $$ BaO $$ is present 
Then $$ Wt. $$ of $$ CaO = (2.5 - x)\,g $$ 

$$ BaO\,\,\,\,\, +\,\,\, H_2SO_4 \rightarrow BaSO_4 \,\,+ \,\, H_2O $$ 
 $$ x\,g $$ 

$$ \dfrac{x}{153} $$ moles    $$ \rightarrow \dfrac{x}{153} $$ moles $$ = \dfrac{x}{153}\times233 = 1.523 x $$ 

 $$ CaO + H_2SO_4 \rightarrow CaSO_4 + H_2O $$ 
 $$ (2.5 - x)\,g $$ 

 $$ \dfrac{(2.5 - x)}{56} $$ moles  $$ \rightarrow \left (\dfrac{2.5 - x}{56}  \right ) $$ moles $$ = \dfrac{(2.5 - x)}{56} \times 136 = (2.5 - x)2.43\,g $$
 
A/q , $$ Wt. $$ of $$ CaSO_4 + Wt. $$ of $$ BaSO_4 = 4.713 \,g $$ 
         $$ (2.5 -x ) 2.43 + 1.523x = 4.713 $$ 

         $$ 6.07 - 2.43x + 1.523 x = 4.713 $$ 

                  $$ 1.358 = 0.907x $$ 
                    $$ x = \dfrac{1.358}{0.907} = 1.497\,g = 1.5\,g $$ 

$$\displaystyle \therefore \, $$ % of $$ BaO = \dfrac{x}{2.5}\times 100 = \dfrac{1.5}{2.5} \times 100 = 60 $$ % 

  


$$4.00g$$ of mixture of $$NaCl$$ and $${Na}_{2}{CO}_{3}$$ was dissolved in water and the volume made up to $$250mL$$; $$25mL$$ of this solution required $$50mL$$ of $$N/10$$ $$HCl$$ for complete neutraliztion. Calculate percentage composition of the original mixture.

$$4g$$ of $$NaCl+{Na}_{2}{CO}_{3}$$ is dissolved in $$250ml$$

$$25ml$$ is taken for reaction,

$${Na}_{2}{CO}_{3}+HCl\rightarrow NaCl+{H}_{2}O+{CO}_{2}$$
$$V=25ml$$    $$50mL$$
$$v.f=2$$        $$\cfrac{1}{10}N$$

Mol.wt $$=106$$

meq. of $${Na}_{2}{CO}_{3}=$$ meq of $$HCl$$ used

$$\cfrac{wt\quad in\quad 25ml}{106/2}\times 1000=50\times \cfrac{1}{10}$$
wt in $$25ml=\cfrac{5\times 53}{1000}$$

Wt in $$250ml=\cfrac{5\times 53}{100}\times \cfrac{250}{25}=2.65g$$

% of $${Na}_{2}{CO}_{3}=\cfrac{2.65}{4}\times 100=66.25$$%
% of $$NaCl=\cfrac{1.35}{4}\times 100=33.75$$%


A mixture of $$0.535$$ g of ethanol and acetaldehyde when heated with Fehling’s solution gave $$1.2$$ g of a red precipitate. What is the percentage of acetaldehyde in the mixture? (Cu = $$63.5$$)

$$W(C_2H_5OH + CH_3CHO) = 0.535\ g$$

Let x g of $$CH_3CHO$$ is present in the mixture

$$2CH_3CHO + 2Fehling \space solution \rightarrow 2CH_3COOH + Cu_2O$$
X g                     $$(Cu^{2+})$$                                                 1.2 g

2 mole of $$CH_3CHO = 1$$ mole of $$Cu_2O$$

$$\Rightarrow$$ moles of $$CH_3CHO = \dfrac{1}{2}$$ (moles of $$Cu_2O$$ produced) $$= \dfrac{1}{2} \left(\dfrac{1.2}{63.5 \times 2 + 16}\right) = \dfrac{1}{2} \left(\dfrac{1.2}{143}\right)$$ 

$$\dfrac{Wt_{CH_3CHO}}{Mol. wt} = 8.39 \times 10^{-3} \times \dfrac{1}{2}$$

$$Wt_{ CH_3CHO} = \dfrac{1}{2} \times 8.39 \times 10^{-3} \times (24 + 4 + 16) = \dfrac{0.369}{2} g = 0.1845 g$$

$$\therefore$$ % of $$CH_2CHO = \dfrac{0.1845}{0.535} \times 100 = 34.5\%$$


A mixture of $$KMn{O}_{4}$$ and $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$$ weighing $$0.24g$$ on being treated with $$KI$$ in acid solution liberates just sufficient $${I}_{2}$$ to react with $$60mL$$ of $$0.1N$$ $${Na}_{2}{S}_{2}{O}_{3}$$ solution. Calculate percentage of $$Cr$$ and $$Mn$$ in the mixture.

$$KMn{O}_{4}+{K}_{2}{Cr}_{2}{O}_{7}+KI\xrightarrow [  ]{ { H }^{ + } } {I}_{2}+{Cr}^{3+}+{Mn}^{2+}$$
$$v.f=5$$  $$v.f=6$$      $$v.f=1$$   $$v.f=2$$

Also $${I}_{2}+hypo\rightarrow {I}^{-}+{S}_{4}{O}_{6}^{2-}$$
$$(v.f=2)$$  $$({Na}_{2}{S}_{2}{O}_{3}$$)

meq of $$KMn{O}_{4}+$$ meq of $${K}_{2}{Cr}_{2}{O}_{7}=$$ meq of $${I}_{2}=$$ meq of hypo solution

                                                                  $$=60\times 0.1=6 \ meq$$

mmole of $$KMn{O}_{4}\times 5+$$ mmole of $${K}_{2}{Cr}_{2}{O}_{7}\times 6=6$$

$$\cfrac{wt.KMn{O}_{4}}{158}\times 5\times 1000+\cfrac{(0.24-{wt}_{KMn{O}_{4}})}{294}\times 6\times 1000=6$$

Wt of $$KMn{O}_{4}=x$$ (say)

Wt of $${K}_{2}{Cr}_{2}{O}_{7}=(0.24-x)y$$

$$\cfrac{5x}{158}+\cfrac{6(0.24-x)}{294}=\cfrac{6}{1000}$$

$$\cfrac{1470x-948x+227.52}{158\times 294}=\cfrac{6}{1000}$$

$$552x=278.712-227.52=52.192$$

$$x=\cfrac{51.192}{522}=0.098$$

wt of $$KMn{O}_{4}=0.098g$$

wt of $${K}_{2}{Cr}_{2}{O}_{7}=0.142g$$

wt of $$Cr=\cfrac{0.142}{294}\times 2\times 52=0.05$$

wt of $$Mn=\cfrac{0.098}{158}\times 55=0.034$$

% of $$Cr=\cfrac{0.05}{0.24}\times 100=20.8$$%

% of $$Mn=\cfrac{0.034}{0.24}\times 100=14.17$$%


A $$20.0 \,cm^{3}$$ mixture of $$CO, CH_{4}$$ and $$He$$ gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be $$13.0 \,cm^{3}.$$ A further contraction of $$14.0\,cm^{3}$$ occurs when the residual gas is treated with $$KOH$$ solution. Find out the composition of the gases mixture.

Let the vol. of $$CO$$ in the mixture $$= x \,cm^{3}$$ and that of $$CH_{4} = y \,cm^{3}$$

$$\therefore$$ Vol. of He in the mixture $$= 20 - (x + y) \,cm^{3}$$

Contraction in vol. due to explosion and cooling $$= 13.0 \,cm^{3}$$ and contraction in volume due to absorption in $$KOH = 14.0 \,cm^{3} =$$ vol. of $$CO_{2}$$ produced.

The chemical equations representing combustion of $$CO$$ and $$CH_{4}$$ respectively are:

$$\underset{x \,cm^{3}}{2CO} + \underset{x/2 \,cm^3}{O_{2}} \rightarrow \underset{x\,cm^3}{2CO_2}$$ ..........(i)

$$\underset{y\,cm^3}{CH_{4}} + \underset{2y\,cm^{3}}{2O_2} \rightarrow \underset{y \,cm^3}{CO_2} + \underset{neg. \,volume}{2H_2O}$$ .............(ii)

Contraction in volume due to reaction (i)

$$x + \dfrac{x}{2} - x = \dfrac{x}{2} \,cm^{3}$$

Contraction in volume due to reaction (ii)

$$y + 2y - y = 2 \,y \,cm^{3}$$

$$\therefore$$ Total contraction in volume

$$= \left ( \dfrac{x}{2} + 2y \right ) cm^{3}$$

But $$\dfrac{x}{2} + 2y = 13$$ (Given) .........(iii)

Total volume of $$CO_{2}$$ produced in reactions (i) and (ii)

$$= (x + y) \,cm^{3}$$

But $$x + y = 14$$ (Given) ...........(iv)

Solving (iii) and (iv) for x and y, we have,

$$x = 10.0 \,cm^{3}$$ and $$y = 4.0 \,cm^{3}$$

Thus, volume of $$CO = 10 \,cm^{3},$$ of $$CH_{4} = 4 \,cm^{3}$$ and of He $$= 20 - (10 + 4) = 6 \,cm^{3}.$$


A mixture of $$ \mathrm{FeO} $$ and $$ \mathrm{Fe}_{3} \mathrm{O}_{4} $$ was heated in air to a constant mass. It was found to gain $$ 10 \% $$ in its mass. Calculate the percentage composition of the original mixture.

On heating in air, both $$ \mathrm{FeO} $$ and $$ \mathrm{Fe}_{3} \mathrm{O}_{4} $$ are oxidized to $$ \mathrm{Fe}_{2} \mathrm{O}_{3} $$ as follows:
$$\underset{2(56+16)}{2FeO}+\dfrac{1}{2}O_2 \rightarrow \underset{2 \times 56+3 \times 16)}{Fe_2O_3}$$
$$=144 g \qquad \qquad =160g$$
$$\underset{2(3 \times 56+4 \times16)}{2Fe_3O_4}+\dfrac{1}{2}O_2 \rightarrow \underset{3 \times 160}{3Fe_2O_3}$$
$$=464 g \qquad \qquad \qquad =480g$$

Suppose $$ \mathrm{FeO} $$ in the mixture $$ =x \% $$
Then $$ \mathrm{Fe}_{3} \mathrm{O}_{4} $$ in the mixture $$ =(100-x) \% $$
$$ \mathrm{Fe}_{2} \mathrm{O}_{3} $$ produced from $$ x g $$ of $$ \mathrm{FeO}=\dfrac{160}{144} \times x g $$
$$ \mathrm{Fe}_{2} \mathrm{O}_{3} $$ produced from $$ (100-x) g $$ of $$ \mathrm{Fe}_{3} \mathrm{O}_{4} $$
$$=\dfrac{480}{464} \times(100-x) g$$
Total $$ \mathrm{Fe}_{2} \mathrm{O}_{3} $$ produced from $$ 100 \mathrm{g} $$ of the mixture $$ =100+10=110 \mathrm{g} $$
$$\therefore \quad \dfrac{160 x}{144}+\dfrac{480(100-x)}{464}=110$$
or $$ \quad \dfrac{10 x}{9}+\dfrac{30\left(\dfrac{100}{29}-x\right)}{29}=110 $$
or $$ \quad 290 x+270(100-x)=110 \times 9 \times 29 $$
or $$ \quad 20 x=1710 $$
or $$ x=85 \cdot 5 $$
$$ \therefore \mathrm{FeO} $$ present in the mixture $$ =85 \cdot 5 \% $$
and $$ \quad \mathrm{Fe}_{3} \mathrm{O}_{4}=100-85 \cdot 5=14 \cdot 5 \% $$


Class 11 Engineering Chemistry Extra Questions