Question 15 Marks
A gaseous compound of nitrogen and hydrogen contains $12.5\%$ hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is $37.$
AnswerCalculation of empirical formula:
Percentage of hydrogen $= 12.5\%$
Percentage of nitrogen $= 100-12.5 = 87.5\%$
| Element |
Atomic mass |
Percentage |
Relative Number of moles |
Simplest mole ratio |
Whole number ratio |
| $N$ |
$14$ |
$87.5$ |
$87.5/14 =6.25$ |
$6.25/6.25 =1$ |
$1$ |
| $H$ |
$1$ |
$12.5$ |
$12.5/1 =12.5$ |
$12.5/6.25 =2$ |
$2$ |
Empirical formula $= NH_2$
Given molecular mass $= 37$
Empirical formula mass $= 16$
$n =$ Molecular mass/Empirica l Formula mass
$= 37/16 = 2.3$ or approximately $2$
Molecular formula $= n \ \times$ Empirical formula
$= 2 \times NH_2$
$= N_2H_4$ View full question & answer→Question 25 Marks
$(i)$ A compound has the following percentage composition by mass $: C$ is $14\%, H$ is $1.2\%$ and $Cl$ is $84.5\%$
Determine the empirical formula of this compound.
$(ii)$ The relative molecular mass of this compound is $168,$ so what is its molecular formula?
$(iii)$ By what type of reaction could this compound be obtained from ethylene?
Answer
| Element |
Atomic mass |
Percentage |
Relative Number of moles |
Simplest mole ratio |
Whole number ratio |
| $C$ |
$12$ |
$14.4$ |
$14.4/12 =1.2$ |
$1.2/1.2 = 1$ |
$1$ |
| $H$ |
$1$ |
$1.2$ |
$1.2/1 = 1.2$ |
$1.2/1.2 = 1$ |
$1 $ |
| $Cl$ |
$35.5$ |
$84.5$ |
$84.5/35.5 =2.4$ |
$2.4/1.2 = 2$ |
$2$ |
Empirical formula of the compound is $CHCl_{2.}$
$(ii)$ Now, empirical formula mass $= 12 + 1 + 35.5 \times 2$
$= 12 + 1 + 71 = 84$
$n =$ Relative molecular mass/ Empirical Formula mass
$= 168 / 84 = 2$
So, molecular formula $= n \times ($Empirical Formula$)$
$= 2 \times CHCl_2$
$= C_2H_2Cl_{4.}$
$(iii)$ Addition reaction with chlorine. View full question & answer→Question 35 Marks
The reaction of potassium permanganate $(VII)$ with acidified iron $(II)$ sulphate is given below :
$2KMno_4 + 10FeSO_4 + 8H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 5Fe2(SO_4)_3 + 8H_2O$
If $15.8g$ of potassium permanganate $(VII)$ was used in the reaction, calculate the mass of iron $(II)$ sulphate used in the above reaction.
AnswerMolecular weight of $KMnO_4 = 39 + 55 + 16 \times 4 = 158$
Molecular weight of $K_2SO_4 = 2 \times 39 + 32 + 16 \times 4 = 174$
Molecular weight of $FeSO_4 = 56 + 32 + 64 = 152$
$2 x 158g$ of $KMnO_4$ yields $= 174g$ of $K_2SO_4.$
So$, 15.8g$ of $KMnO_4$ will yield $= 174 \times 15.8/2 \times 158 = 8.7g$ of $K_2SO_4 .$
$174g$ of $K_2SO_4$ yields $152g$ of $FeSO_4$
So$, 8.7 g$ of $K_2SO_4$ will yield $= 152 \times 8.7/174 = 7.6 g$ of $FeSO_{4.}$
Hence$, 7.6g$ of iron $(II)$ sulphate is used in the above reaction.
View full question & answer→Question 45 Marks
The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular mass. Given $8 g$ of each gas at $\text{STP},$ which gas will contain the least number of molecules and which gas the most?
AnswerCalculation of number of molecules in each gas sample :
$(i) \ 8g$ of hydrogen :
$1g$ of hydrogen $= 6.023 \times 10^{23}$ molecules
$8 g$ of hydrogen will have $= 6 .023 \times 1023 \times 8 = 48 .184 \times 1023$ molecules
$(ii ) \ 8 g$ of oxygen :
$32 g$ of oxygen has $= 6 .023 \times 1023$ molecules
$8 g$ of oxygen will have $= 6 .023 \times 1023/ 32 \times 8 = 1.50 \times 1023$ molecules
$(iii ) \ 8 g$ of carbon dioxide :
$44 g$ of carbon dioxide has $= 6 .023 \times 1023$ molecules
$8 g$ of carbon dioxide will have $= 6 .023 \times 1023/ 44 \times 8 = 1.0 \times 1023$ molecules
$(iv)\ 8$ g of sulphur dioxide :
$64$ of sulphur dioxide has $= 6 .023 \times 1023$ molecules
$8 g$ of sulphur dioxide will have $= 6 .023 \times 1023/ 64 \times 8 = O. 75 \times l 023$
molecules
$(v)\ 8 g$ of chlorine :
$35 .5 g$ of chlorine has $= 6 .023 \times 1023$ molecules
$8 g$ of chlorine will have $= 6 .023 \times 1023/ 71 \times 8 = 0 .68 \times 1023$ molecules
Hence, chlorine gas will have least number of molecules. Hydrogen gas will have
the most number of molecules.
View full question & answer→Question 55 Marks
A metal $M,$ forms a volatile chloride containing $65.5\%$ Chlorine. If the density of the chloride relative to hydrogen is $162.5,$ find the molecular formula of the chloride. $[M = 56, Cl = 35.5]$
AnswerGiven, density of chloride relative to hydrogen $= 162.5$
Percentage of chlorine $= 65.5\%$
Percentage of Metal $M = 100 - 65.5 = 34.5\%$
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest mole ratio |
Whole number ratio |
| $M$ |
$56$ |
$34.5$ |
$34.5/56 = 0.62$ |
$0.62/0.62 = 1$ |
$1$ |
| $Cl$ |
$35.5$ |
$65.5$ |
$65.5/35.5 =1.85$ |
$1.85/0.62 = 2.98$ |
$3$ |
Empirical formula $= \text{MCl}_3$
Empirical formula mass $= 56 + 3 \times 35.5 = 162.5$
Molecular mass $= 2\ \times$ Vapour density $= 2 \times 162.5 = 325$
$n =$ Molecular mass / Empirical formula mass
$= 325/162.5 = 2$
Molecular formula $= n\ \times$ empirical formula
$= 2 \times MCl_{3 }= M_2Cl_6$ View full question & answer→Question 65 Marks
Chlorine, nitrogen, ammonia and sulphur dioxide gases are collected under the same conditions of temperature and pressure.
Copy the following table which gives the volumes of the gases collected, and the number of molecules (X) in 20L of nitrogen.You are to complete the table by giving the number of molecules in th e other gases, in terms of X.
| Gas | Volume(litres) | Number of molecules |
| Chlorine | 10 | |
| Nitrogen | 20 | X |
| Ammonia | 20 | |
| Sulphur dioxide | 5 | |
Answer(a) If 20L of nitrogen has = X number of molecules
Then, 10L of chlorine will have = X x 10/20 = X/2.
(b) If 20L of nitrogen has = X number of molecules
Then, 20L of ammonia will have = X x 20/20 = X.
(c) If 20L of nitrogen has = X number of molecules
Then, 5L of sulphur dioxide will have = X x 5/20 = X/4.
| Gas | Volume(litres) | Number of molecules |
| Chlorine | 10 | X/2 |
| Nitrogen | 20 | X |
| Ammonia | 20 | X |
| Sulphur dioxide | 5 | X/4 |
View full question & answer→Question 75 Marks
The percentage composition of sodium phosphate, as determined by analysis is $: 42.1\% \ Na, 18.9\%\ P, 39\%$ of $O$. Find the empirical formula of the compound.
$[H =1, N =14, Na = 23, P = 31, Cl = 35.5, Pt = 195]$
Answer
| Element |
Atomic mass |
Percentage |
Relative ratio of moles |
Simplest mole ratio |
Whole number ratio |
| $Na$ |
$23$ |
$42.1$ |
$42.1/23 = 1.8$ |
$1.8/0.6 =3$ |
$3$ |
| $P$ |
$31$ |
$18.9$ |
$18.9/31 = 0.6$ |
$0.6/0.6
= 1$ |
$1 $ |
| $O$ |
$16$ |
$39$ |
$39/16
= 2.4$ |
$2.4/0.6 = 4$ |
$4$ |
Empirical formula $= Na_3PO_4.$ View full question & answer→Question 85 Marks
Determine the empirical formula of the compound whose composition by mass is $42\%$ nitrogen$, 48\%$ oxygen and $9\%$ hydrogen.$\ [N=14,O=16,H=1]$
Answer
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest mole ratio |
Whole number ratio |
| $N$ |
$14$ |
$42$ |
$42/14 = 3$ |
$3/3 = 1$ |
$1$ |
| $O$ |
$16$ |
$48$ |
$48/16 = 3$ |
$3/3 = 1$ |
$1$ |
| $H$ |
$1$ |
$9$ |
$9/1 = 9$ |
$9/3 = 3$ |
$3$ |
So the empirical formula of the compound is $NH_2OH.$ View full question & answer→Question 95 Marks
Concentrated nitric acid oxidizes phosphorous to phosphoric acid according to the following equation :
$P + 5HNO_3 \rightarrow H_3PO_4 + H_2O + 5NO_2$
What will be the volume of steam at the same time measured at $760\ mm$ of Hg pressure and $273^\circ C$ ? $[H=1, N=14, O=16, P=31]$
AnswerGiven equation is :
$ P +5 HNO _3 \rightarrow H _3 PO _4+ H _2 O +5 NO _2 $
Moles of steam formed from $31 g$ phosphorus $=1 mol$
moles of steam from $6.2 g$ phosphorus $=1 mol \times 6.2 g / 31 g$
$=0.2 mol$.
volume of steam produced at $S.T.P =(0.2 \ mol ) \times(22.4 L / mol )$
$=4.48$ litre.
Since the pressure $(760\ mm )$ remains constant, but the temperature $(273+273)=546$ is doubled, the volume of the steam also gets doubled
$\therefore$ volume of steam produced at $760\ mm\ Hg$ and $273^{\circ} C$
$ =4.48 \times 2=8.96 \text { litres. } $
View full question & answer→Question 105 Marks
The compound A has the following percentage composition by mass: $C =26.7\%, O = 71.1\%, H = 2.2\%.$
Determine the empirical formula of $A$.
$($Answer to one decimal place$) \ (H=1,C=12,O=16)$
Answer
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest mole ratio |
Whole number ratio |
| $C$ |
$12$ |
$26.7$ |
$26.7/12 = 2.2$ |
$2.2/2.2 = 1$ |
$1$ |
| $O$ |
$16$ |
$71.1$ |
$71.1/16 = 4.44$ |
$4.44/2.2 =2$ |
$2$ |
| $H$ |
$1$ |
$2.2$ |
$2.2/1
= 2.2$ |
$2.2/2.2 = 1$ |
$1$ |
So the empirical formula of the compound is $CO_2H.$ View full question & answer→Question 115 Marks
Calculate the empirical formula of the compound having $37.6\%$ sodium$, 23.1\%$ silicon and $39.3\%$ oxygen.$($Answer correct to two decimal places$) (O = 16, Na = 23, Si = 28)$
Answer
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest mole ratio |
Whole number ratio |
| $Na$ |
$23$ |
$37.6$ |
$37.6/23 = 1.63$ |
$1.63/0.83 = 1.9$ |
$2$ |
| $Si$ |
$28$ |
$23.1$ |
$23.1/28 = 0.83$ |
$0.83/0.83 = 1$ |
$1$ |
| $O$ |
$16$ |
$39.3$ |
$39.3/16 = 2.45$ |
$2.45/0.83 = 2.9$ |
$3$ |
So the empirical formula of the compound is $Na_2SiO_3.$ View full question & answer→Question 125 Marks
A compound contains $87.5\%$ by mass of nitrogen and $12.5\%$ by mass of hydrogen. Determine the empirical formula of this compound.
AnswerEmpirical formula of the compound is as :
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest mole ratio |
Whole number ratio |
| $ N$ |
$14$ |
$87.5$ |
$87.5/14 = 6.25$ |
$6.25/6.25 = 1$ |
$1$ |
| $H$ |
$1$ |
$12.5$ |
$12.5/1
=12.5$ |
$12.5/6.25 = 2$ |
$2$ |
So, the empirical formula of the compound is $NH_2.$ View full question & answer→Question 135 Marks
Under the same conditions of temperature and pressure you collect $2L$ of carbon dioxide$, 3L$ of chlorine$, 5L$ of hydrogen$, 4L$ of nitrogen and $1L$ of sulphur dioxide. In which gas sample will there be :
$a.$ The greatest number of molecules.
$b.$ The least number of molecules.
Justify your answer.
AnswerCalculation of number of molecules in each gas sample:
$(i)\ 2L$ of carbon dioxide :
$22.4L$ of carbon dioxide has $= 6.023 \times 10^{23}$ molecules
$2L$ of carbon dioxide will have $= 6.023 \times 10^{23} \times 2/22.4$
$= 0.5377 \times 10^{23}$ molecules
$(ii )\ 3L$ of chlorine :
$22.4L$ of chlorine has $= 6.023 \times 10^{23}$ molecules
$3L$ of chlorine will have $= 6.023 \times 10^{23}\times 3/22.4$
$= 0.8066 \times10^{23}$ molecules
$(iii ) \ 5L$ of hydrogen :
$22.4L$ of hydrogen has $= 6.023 \times 10^{23}$ molecules
$5L$ of hydrogen will have $= 6.023 \times 10^{23} \times 5/ 22.4$
$= 1.34 \times10^{23}$ molecules
$(iv) \ 4L$ of nitrogen :
$22 .4 L$ of nitrogen has $= 6.023 \times 10^{23}$ molecules
$4 L$ of nitrogen will have $= 6.023 \times 10^{23} \times 4/ 22 .4$
$= 1.07 \times 10^{23}$ molecules
$(v)\ 1L$ of sulphur dioxide :
$22.4L$ of sulphur dioxide has $= 6.023 \times 10^{23}$ molecules
$1 L$ of sulphur dioxide will have $= 6.023 \times 10^{23}\times 1/22.4$
$= 0.27 \times 10^{23}$ molecules
From the above calculation of number of molecules in different gases we can conclude that the :
$(a)$ The greatest number of molecules are present in $5L$ of hydrogen gas sample.
$(b )$ The least number of molecules is in $1L$ of sulphur dioxide gas sample.
View full question & answer→Question 145 Marks
A compound of lead has following percentage composition$, Pb = 90.66\%, O = 9.34\%.$ Calculate empirical formula of a compound. $[Pb = 207, O = 16]$
AnswerEmpirical formula of a compound:
| Element |
Atomic
mass |
Percentage |
Relative number of |
Simplest mole ratio |
Whole number ratio |
| $Pb$ |
$207$ |
$90.66$ |
$90.66/207 = 0.44$ |
$0.44/0.44 = 1$ |
$1 \times 3 = 3$ |
| $O$ |
$16$ |
$9.34$ |
$9.34/16 = 0.58$ |
$0.58/0.44 = 1.32$ |
$1.32 \times 3 = 3.96 = 4$ |
Since the mole ratio for oxygen is fractional so we multiply the whole ratio by $3$ to make it a whole number. So, the empirical formula of the compound is $Pb_3O_4.$ View full question & answer→Question 155 Marks
The usefulness of a fertilizer depends upon percentage of nitrogen present in it. Find which of the following is a better fertilizer:
$(a)$ Ammonium nitrate $[NH_4NO_3]$
$(b)$ Ammonium phosphate $[(NH_4)_3PO_4(N=14,H=1,O=16,P=31)]$
Answer$(a)$ Percentage of nitrogen in Ammonium nitrate $[NH_4NO_3] :$
$(N)_2 + (H)_4 + (O)_3$
$14 \times 2 + 1 \times 4 + 3 \times 16 = 80.$
So, the molecular mass of $NH_4NO_3 = 80.$
$80$ by weight of $NH_4NO_3$ contain $28$ parts by weight of nitrogen .
$100$ parts will contain $= 28 \times 100 /80 = 35\%$
So, the percentage composition of nitrogen in Ammonium nitrate is $3535\%$
$(b)$ Percentage of nitrogen in Ammonium phosphate $[(NH_4)_3PO_4] :$
$(N)_3 + (H)_{12} +P + (O)_4$
$14 \times 3 + 1 \times 12+ 31 + 16 \times 4 = 149 .$
So, the molecular mass of $NH_4NO_3 = 149 .$
$149$ by weight of $(NH_4)_3PO_4.$ contain $42$ parts by weight of nitrogen .
$100$ parts will contain $= 42 \times 100 /149 = 28.18\%$
So, the percentage of nitrogen in ammonium phosphate is $28.18\%$
Since the percentage of nitrogen is more in Ammonium nitrate so it is a better fertilizer.
View full question & answer→Question 165 Marks
Calculate the volume of air at $\text{STP},$ required to convert $300 \ mL$ of sulphur dioxide to sulphur trioxide. Air contains $21\%$ of oxygen by volume.
AnswerConversion of sulphur dioxide to sulphur trioxide follows the following balanced equation :
$4SO_2 + 2O_2 \rightarrow 4SO_3$
According to Gay -lussac's law:
$4\ vol. 2\ vol. 4\ vol.$
Volume of $SO_2 = 4\ vol.$
$=300\ ml$
Volume of $0 2 = 2\ vol.$
$=300 \times 2 / 4$
$= 150\ ml$
Volume of oxygen required $= 21 ^\circ$
$= 150 \ ml$
Volume of air required at $\text{STP} = 100^\circ$
$= 100 \times 150 / 21$
$= 714.28 \ ml$
So, the volume of air at $\text{STP},$ required to convert $300\ ml$ of sulphur dioxide to
sulphur trioxide is $714.28\ ml$
View full question & answer→Question 175 Marks
Two oxides of a metal $(M)$ have $20.12\%$ and $11.19\%$ oxygen. The formula of the first oxide is $MO$. Determine the formula of the second oxide.
AnswerCalculation of molar mass of $M$ from first oxide :
Let us assume the atomic mass of Mas $x$.
Atomic mass of Oxygen $= 16$
| Element |
Percentage |
Relative number of moles |
Simplest mole ratio |
| $M$ |
$100 - 20.12 = 79.88$ |
$79.88/x$ |
$79.88/1.25x$ |
| $O$ |
$20.12$ |
$20.12/16=1.25$ |
$1.25/1.25 = 1$ |
So, molar mass of $M, x = 63.5$
Calculation Of formula of second oxide :
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest
mole ratio |
Whole number ratio |
| $M$ |
$63.5$ |
$88.81$ |
$88.81/63.5 = 1.4$ |
$1.4/0.69 = 2$ |
$2$ |
| $O$ |
$16$ |
$11.19$ |
$11.19/16 = 0.69$ |
$0.69/0.69 =1$ |
$1$ |
So formula of second oxide is $M_2O.$ View full question & answer→Question 185 Marks
An organic compound has the following percentage composition $: C = 12.76\%, H = 2.13\%, Br = 85.11\%$. The vapour density of the compound is $94.$ Find out its molecular formula.
Answer
| Element |
Atomic mass |
Percentage |
Relative number of moles |
Simplest
mole ratio |
Whole number ratio |
| $C$ |
$12$ |
$12.76$ |
$12.76/12 = 1.06$ |
$1.06/1.06 = 1$ |
$1$ |
| $H$ |
$1$ |
$2.13$ |
$2.13/1 =2.13$ |
$2.13/1.06 = 2$ |
$2$ |
| $Br$ |
$80$ |
$85.11$ |
$85.11/80 =1.06$ |
$1.06/1.06 = 1$ |
$1$ |
So the Empirical formula of the compound will be $CH_2Br.$
Now the Empirical formula mass will be $=$ Atomic mass of $C\ +$ Atomic mass of $H\ +$ Atomic mass of $Br$
$= 12 + 1 \times 2 + 80 = 94$
Now as Molecular mass $= 2\ \times$ vapour density
$= 2 \times 94 = 188.$
So n = Molecular mass / empirical formula mass
$= 188/ 94 = 2$
Molecular formula of the compound is $= n \ \times$ empirical formula
$= 2 \times (CH_2Br) = C_2H_4Br_2$ View full question & answer→Question 195 Marks
Calculate the percentage of water in ferrous sulphate crystals.
$[Fe = 56, S = 32, O =16, H = 1].$
AnswerFerrous sulphate is $FeSO_4.7H_2O$
Molecular mass of Ferrous sulphate is $FeSO_4.7H_2O$ is:
Atomic mass of $Fe \ +$ Atomic mass of $S \ +$ Atomic mass of $H \ +$ Atomic mass of $O$
$= 56 + 32 + 1 \times 14 + 16 \times 11 = 278$
$278$ parts by weight of crysta ls contain $126$ parts of water
$100$ parts will contain $= 126 \times 100 / 278 = 45.32\%$
So, the percentage of water in ferrous sulphate crystals is $45.32\%$
View full question & answer→Question 205 Marks
Explain the term $: Mole$
AnswerThe $mole$ is a unit of measurement for the amount of substance. Moles give us a consistent method to convert between atoms/molecules and grams.
The mathematical expression for calculating moles is expressed as:
$moles =$ mass / molar mass
We can express moles in terms of molecular mass, number of atoms/molecules and in terms of volume.
- $1\ mole$ of a pure substance has a mass in grams equal to its molecular mass.
- $1 \ mole$ contains the same number of particles as there are in $12g$ of carbon-$12$ atoms. This number is called Avogadro’s number and is equal to $6.023 \times 10^{23}$ particles.$1 \ mole$ of a gas occupies a volume of: $22.4$ litres at $\text{S.T.P}.$
For example: $1 \ mole$ of sodium atom has mass $23 g$. It contains $6.023 \times 10^{23}$ atoms and occupies $22.4$ litres of volume at $\text{S.T.P}.$ View full question & answer→Question 215 Marks
The empirical formula of a compound is $C_2H_5$. Its vapour density is $29$. Determine the relative molecular mass of the compound and hence its molecular formula.
AnswerGiven the empirical formula of the compound is $C_2H_5.$
Vapour density is $= 29 .$
Empirical formula mass of the compound $= 29$
As, molecular mass $= 2\ \times$ vapour density $= 2 \times 29 = 58$
So, molecular mass of the compound $= 58$
Molecular formula $= n \times$ Empirical formula
Now, n $=$ Molecular mass/ Empirical formula mass $= 58/ 29 =2$
Molecular formula $= 2 \times (C_2H_5 ) = C_4H_{10}$
So, molecular formula of compound is $= C_4H_{10}$
View full question & answer→Question 225 Marks
How does Avogadro's law explain Gay $-$ lussac's law of combining volumes?
AnswerExplanation of Gay $-$ Lussac's Law: Gay $-$ Lussac had experimentally determined that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas.
According to Avogadro's law, if :
$1$ volume of hydrogen contains n molecules of the gas then $1$ volume of chlorine also contains n molecules of the gas. Therefore $2$ volume of hydrogen chloride
contain $2n$ molecules of the gas.
$H_2 + Cl_2 \rightarrow 2HCI$
$1 \ vol \ 1\ vol \ 2 \ vol\ ($by Gay $-$ Lussac$)$
$n\ n \ 2n \ ($By Avogadro$)$
but hydrogen and chlorine are diatomic.
So, $2$ atoms $+\ 2$ atoms $\rightarrow 2$ molecules
$1$ atom $+ \ 1$ atom $\rightarrow 1$ molecu le
i .e. $1$ molecule of hydrogen chloride is formed when $1$ atom of hydrogen combines with $1$ atom of chlorine. Thus Avogadro's law explains Gay $-$ Lussac's law of combining volumes.
View full question & answer→Question 235 Marks
What are the main applications of Avogadro's law?
AnswerThe main applications of Avogadro's law are : - Explanation of Gay -Lussac's Law
- Determination of atomicity of gases
- Determination of the molecular formula of a gaseous compound.
- Establishes relationship between the relative vapour density of a gas and its relative molecular mass.
- Establishes the relationship between gram molecular weight and volume of a gas at STP.
View full question & answer→Question 245 Marks
A gas cylinder can hold $1\ \ kg$ of hydrogen at room temperature and pressure.
$(a)$ Find the number of moles of hydrogen present
$(b)$ What weight of $CO_2$ can the cylinder hold under similar conditions of temperature and pressure? $(H = 1 C = 12, O = 16)$
$(c)$ If the number of molecules of hydrogen in the cylinder is $X,$ calculate the number of $CO_2$ molecules in the cylinder with the same conditions of temperature and pressure.
$(d)$ State the law that helped you to arrive at the above result.
Answer$(a)$ Given:
Mass of hydrogen $= 1\ kg$ at $298K$ and $1$ at m pressure
Moles of hydrogen $= ?$
Number of moles of hydrogen $=\frac{\text { Mass of hydrogen }}{\text { Gram atomic mass of hydrogen }}$
$=\frac{1000 g }{1 g }$
$= 1000\ moles$ of hydrogen
$(b)$ Molecular mass of $CO_2 = 12 + 2 \times 16 = 44g$
So,
vapour density $(VD) =\frac{\text { mol. Mass }}{2}=\frac{44}{2}=22$
$\text { V.D. }=\frac{\text { mass of certain amount of } CO _2}{\text { mass of equal volume of hydrogen }}=\frac{ m }{1}$
$22=\frac{ m }{1}$
So, mass of $CO_2 = 22 \ Kg$
$(c)$ According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
$(d)$ So, number of molecules of carbon dioxide in the cylinder $=$ number of molecules of hydrogen in the cylinder $= X$
View full question & answer→Question 255 Marks
Propane burns in air according to the following equation :
$C_3H_{8 }+ 5O_{2 }\rightarrow 3CO_2 + 4H_2O$
What volume of propane is consumed on using $1000 \ cm^3$ of air, considering only $20\%$ of air contains oxygen?
AnswerGiven :
$C_3H_{8 }+ 5O_{2 }\rightarrow 3CO_2 + 4H_2O$
Volume of air $= 1000 \ cm^3$
Percentage of oxygen in air $= 20\%$
From the given information,
$C_3H_{8 }+ 5O_{2 }\rightarrow 3CO_2 + 4H_2O$
$1 \ vol \ 5 \ vols \ 3\ vols \ 4\ vols$
According to Gay$-$Lussac's law,
$1 \ vol.$ of propane consumes $5\ vol.$ of oxygen.
Volume of oxygen $= 1000 \ cm^3 \times 20\% = 200 \ cm^3$
Therefore,
Volume of propane burnt for every $200 \ cm^3$ of oxygen,
$= \frac{1}{5} \times 200 = 400 \ cm^{3}$
$40\ cm^3$ of propane is burnt.
View full question & answer→Question 265 Marks
A gaseous hydrocarbon contains $82.76\%$ of carbon. Given that its vapour density is $29,$ find its molecular formula.
$[C = 12, H = 1]$
Answer$\%$ of carbon $= 82.76\%$
$\%$ of hydrogen $= 100 - 82.76 = 17.24\%$
| Element |
% weight |
Atomic weight |
Relative No. of moles |
Simplest Ratio |
| $C$ |
$82.76$ |
$12$ |
$82.76/12 = 6.89$ |
$6.89/6.8 = 1 \times 2 = 2$ |
| $H$ |
$17.24$ |
$1$ |
$17.24/1 = 17.24$ |
$17.24/6.89 =2.5 \times 2 = 5$ |
Empirical formula $= C_2H_5$
Empirical formula weight $= 2 \times 12 + 1 \times 5 = 24 + 5 = 29$
Vapour density $= 29$
Relative molecular mass $= 29 \times 2 = 58$
$N = N =\frac{\text { Relative molecular mass }}{\text { Empirical weight }}$
$=\frac{58}{29}=2$
Molecular formula $=$ n $\times$ empirical formula
$= 2 \times C_2H_5$
$=C_4H_{10}$ View full question & answer→Question 275 Marks
Consider the following reaction and based on the reaction answer the questions that follow :
$\left( NH _4\right)_2 Cr _2 O _7 \xrightarrow{\text { heat }} N _2( g )+4 H _2 O ( g )+ Cr _2 O _3$
Calculate :
$(i)$ The quantity in moles of $(NH_4)_2Cr_2O_7$ if $63 \ gm$ of $(NH_4)_2Cr_2O_7$ is heated.
$(ii)$ The quantity in moles of nitrogen formed.
$(iii)$ The volume in litres or $dm^3$ of $N_2$ evolved at $\text{STP}$.
$(iv)$ The mass in grams of $Cr_2O_3$ formed at the same time
$[$Atomic masses $: H = 1, Cr=52, N=14]$
Answer$(i)$ Molecular mass of ammonium dichromate
$ =2(14+4)+104+112=252 g $
Number of $moles =\frac{\text { weight }}{\text { Molecular weight }}$
$=\frac{63}{252}=0.25$ moles
$(ii) \ 252 g$ of ammonium dichromate gives $22.4 \ dm ^3$ of $N _2$
$63 g$ of ammonium dichromate gives $=\frac{63 \times 22.4}{252}=5.6 L$
No. of $moles =\frac{\text { Weight }}{\text { Molecular weight }}$
$=\frac{5.6}{22.4}$
$=0.25\ moles$
$(iii)\ 252 g$ of ammonium dichromate gives $22.4 \ dm ^3$ of $N _2$
$63 g$ of ammonium dichromate gives gives $\frac{63 \times 22.4}{252}$
$=5.6 L$
$(iv) \ 252 g$ of ammonium dichromate gives $152 g$ of $CrO _3$.
$63 g$ of ammonium dichromate gives $\frac{63 \times 152}{252}$
$=38 g$.
View full question & answer→Question 285 Marks
$O_2$ is evolved by heating $\text{KCIO}_3$ using $\text{MnO}_2$ as a catalyst.
$2 KCI _3 \xrightarrow{ Mno _2} 2 KCl +3 O _2$
$(i)$ Calculate the mass of $\text{KClO}_3$ required to produce $6.72$ litre of $O_2$ at $\text{STP}$.
$(ii)$ Calculate the number of moles of oxygen present in the above volume and also the number of molecules
$(iii)$ Calculate the volume occupied by $0.01\ mole$ of $CO_2$ at $\text{STP}$
Answer$(i) \ 2 KCI _3 \xrightarrow{ Mno _2} 2 KCl +3 O _2$
$2V\ 2V\ 2V$
$3$ Volumes of Oxygen require $KClO _3=2$ Volumes
So, $1\ Vol$. of oxygen will require $KClO _3=\frac{2}{3}$ volumes
So$, 6.72$ litres of Oxygen will require $KClO _3$
$\frac{2}{3} \times 6.72=4.48$ litres
$22.4$ litres of $KClO _3$ has mass $=122.5 g$
So$, 4.48$ litres of $KClO _3$ will have mass $= \frac{122.5}{22.4} \times 4.48$
$ =24.5 g$
$(ii)\ 22.4$ litres of oxygen $=1\ mole$ So$, 6.72$ litres of oxygen $=\frac{6.72}{22.4}=0.3$ moles
No. of molecules present in $1 \ mole$ of $O _2 = 6.023 \times 10^{23}$
So, no. of molecules present in $0.3 \ mole$ of $O _2$
$=6.023 \times 10^{23} \times 0.3$
$ =1.806 \times 10^{23}$
$(iii)$ Volume occupied by $1 \ mole$ of $CO _2$ at $\text{STP} =22.4$ litres
So,volume occupied by $0.01 \ mole$ of $CO _2$ at $\text{STP} =22.4 \times 0.01=0.224$ litres
View full question & answer→Question 295 Marks
An organic compound with vapour density $= 94$ contains $C = 12.67\%, H = 2.13\%$ and $Br = 85.11\%.$ Find the molecule formula. $[$Atomic mass $C = 12, H = 1, Br = 80]$
Answer
| Element |
Relative atomic mass |
$\%$ Compound |
Atomic ratio |
Simplest ratio |
| $H$ |
$1$ |
$2.13$ |
$2.13/1
= 2.13$ |
$2$ |
| $C$ |
$12$ |
$12.67$ |
$12.67/12 =1.055$ |
$2$ |
| $Br$ |
$80$ |
$85.11$ |
$85.11/80 = 1$ |
$1$ |
Empirical formula $= CH_2Br$
$n($Empirical formula mass of $CH_2Br) =$ Molecular mass $( 2 \times VD)$
$n(12 + 2 + 80) = 94 \times 2$
$n = 2$
molecular formula $=$ Empirical formula $\times 2$
$= (CH_2Br) \times 2$
$= C_2H_4Br_2$ |
View full question & answer→