Question 12 Marks
A certain gas $'X\ '$ occupies a volume of $100 \ cm^3$ at $\text{STP}$ and weights $0.5g.$ Find its relative molecular mass.
AnswerAs at $\text{STP}\ 100 \ cm^{3}$ of gas weights $= 0.5 g$
So, at $\text{STP}\ 22400 \ cm^3$ of gas will weight $= 0.5 \times 22400 / 100 = 112g.$
View full question & answer→Question 22 Marks
A gas cylinder contains $24 \times 10^{24}$ molecules of nitrogen gas. If Avogadro's number is $6 \times 10^{23}$ and the relative atomic mass of nitrogen is $14,$ Calculate : Volume of nitrogen at $ \text{STP}$ in $dm^3.$
AnswerAs $6.023 \times 10^{23}$ molecules of nitrogen occupies $= 22.4 dm^3$ at $ \text{STP}$
$24 \times 10^{23}$ molecules will occupy $= 22.4 \times 24 \times 10^{23}/ 6.023 \times 10^{23}$
$= 896 \ dm^3.$
View full question & answer→Question 32 Marks
A gas cylinder contains $24 \times 10^{24}$ molecules of nitrogen gas. If Avogadro's number is $6 \times 10^{23}$ and the relative atomic mass of nitrogen is $14,$
Calculate : Mass of nitrogen gas in the cylinder.
AnswerMolecular weight of nitrogen $=28$
As $6.023 \times 10^{23}$ molecules of nitrogen weigh $=28$
$24 \times 10^{23}$ molecules will weight $=28 \times 24 \times10^{23} / 6.023 \times 10^{23}$
$=28 \times 40=1120 g$
View full question & answer→Question 42 Marks
What is the empirical formula of octane?
AnswerMolecular formula of octane $= C_8H_{18}$
Its empirical formula will be $= C_9H_9$
View full question & answer→Question 52 Marks
The equation for the burning of octane is :
$2C_8H_{18} + 25O_{2 }\rightarrow 16CO_2 + 18H_2O$
If the relative molecular mass of carbon dioxide is $44,$ what is the mass of carbon dioxide produced by burning two $moles$ of octane?
AnswerSince $2\ moles$ of $C_8H_{18}$ produces $= 16 \ moles$ of $CO_2$
$= 16 \times 44 = 704$
So by burning $2 \ moles$ of octane $704g$ of $CO_2$ is produced.
View full question & answer→Question 62 Marks
The equation for the burning of octane is :
$2C_8H_{18} + 25O_2\rightarrow 16CO_2 + 18H_2O$
What volume, at $\text{STP},$ is occupied by $8 \ moles$ ?
AnswerNow$, 1 \ mole$ of $CO_2$ occupies $= 22.4L$ at $\text{STP}.$
So$, 8 \ moles$ of $CO_2$ will occupy $= 22.4 \times 8 = 179.2L$ at $\text{STP}.$
View full question & answer→Question 72 Marks
A sample of ammonium nitrate when heated yields $8.96$ litres of steam $($measured at $\text{STP})$
$NH_4NO_3 \rightarrow N_2O + 2H_2O$
Determine the percentage of oxygen in the ammonium nitrate?
Answer$80$ parts by weight of $NH_4NO_3$ contains $48$ parts by weight of oxygen
So $100$ parts will contain $= 48 \times 100/80 = 60\%$
View full question & answer→Question 82 Marks
A sample of ammonium nitrate when heated yields $8.96$ litres of steam $($measured at $\text{STP})$
$NH_4NO_3 \rightarrow N_2O + 2H_2O$
What mass of ammonium nitrate should be heated to produce $8.96L$ of steam?
AnswerMolecular mass of $NH_4NO_3 = 80$
$44.8L$ of steam is liberated by $= 80g$ of $NH_4NO_3$
$8.96L$ of steam will be liberated by $= 80 \times 8.96 / 44.8 = 16L$
View full question & answer→Question 92 Marks
Calculate the percentage of sodium in sodium aluminium fluoride $(Na_3AIF_6).$
$[F = 19, Na = 23, Al = 27]$
AnswerMolecular weight of sodium aluminium fluoride $(Na_3AIF_6 ) = 210$
Now$, 210$ parts by weight of $Na_3AIF_6$ contains $= 69$ parts by weight of sodium
So$, 100$ parts will contain $= 69 \times 100 / 210 = 32.8$ or $33\%$
View full question & answer→Question 102 Marks
The volumes of gases $A, B, C$ and $D$ are in the ratio$, 1:2:2:4$ under the same conditions of temperature and pressure.
The number of molecules in the actual volume of $D$ at $\text{STP}$ is $6 \times 10^{23}$ molecules. state, the mass of $D$ if the gas is dinitrogen oxide $(N_2O)$
Answer$6 \times 10^{23}$ molecules is Avogadro's number of molecules contained in one gram mole of the substance if gas $D$ is $N _2 O$
then$, 1$ gram $mole$ of $N _2 O =2 \times 14+16=44 g$
View full question & answer→Question 112 Marks
The equations given below relate to the manufacture of sodium carbonate $($Molecular weight of $Na_2CO_3 = 106).$
$NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl$
$2NaHCO_3\rightarrow Na_2CO_3 + H_2O + CO_2$
To produce the mass of sodium hydrogen carbonate what volume of carbon dioxide, measured at $\text{STP}$ would be required?based on the production of $21.2 g$ of sodium carbonate.
Answer$NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl $
From the equation$, 1\ mole$ of $CO_2$ i.e. $22.4L$ of $CO_2$ is used to produce $= 1\ mole$ of $NaHCO_3$
Now as$, 84g$ of $NaHCO_3$_ requires $= 22.4L$ of $CO_2$
$33.6 g$ of $NaHCO_3$ will require $= 22.4 \times 33.6/84 = 8.96L$ of $CO_2.$
View full question & answer→Question 122 Marks
Samples of the gases $O_2, N_2, CO_2$ and $CO$ under the same conditions of temperature and pressure contain the same number of molecules represented by $X$. The molecules of Oxygen, occupy $V$ litres and have a mass of $8 g$. Under the same conditions of temperature and pressure :
What is the mass of $CO_2$ in $g$?
Answer$X$ molecules of $O_2 = 8/32$
$= 1/4 \ mole$ of $O_2$
So$, X$ molecules of $CO_2 = 1/4$ molecule of $CO_2$
So, Mass of $CO_2$ present in the sample $= 1/4\ \times$ gram molecular mass of $CO_2$
$= 1/4 \times 44 = 11g$
View full question & answer→Question 132 Marks
Calculate the percentage of phosphorous in the fertilizer superphosphate$, Ca(H_2PO_4)_2. [Ca = 40, H =1, P =31, O = 16]\ ($Correct to $1$ decimal place$)$
AnswerMolecular mass of fertilizer superphosphate$, Ca(H_2PO_4)_2 =234$
$234$ parts by weight of fertilizer contains $62$ parts by weight of phosphorous
So$, 100$ parts will contain $= 62 \times 100/234 = 26.5\%$
View full question & answer→Question 142 Marks
If $112 \ cm^3$ of hydrogen sulphide is mixed with $120 \ cm^3$ of chlorine at $\text{STP},$ what is the mass of sulphur formed?
$H_2S + Cl_2 \rightarrow 2HCI + S$
AnswerAt $\text{STP}, 22400\ cm^3$ of each $H_2S$ and $Cl_2$ will give $32g$ of sulphur i.e.
$44800 \ cm^3$ of $H_2S + Cl_2$ gives $= 32g$ of $S$
$(112 +120) = 232\ cm3$ of $H_2S + Cl_2$ will give $= 32 \times 232 / 44800 = 0.16g$ of $S$
View full question & answer→Question 152 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 mass of nitric acid will be consumed at the same time?
AnswerGiven equation is :
$P +5 HNO _3 \rightarrow H _3 PO _4+ H _2 O +5 NO _2$
Molecular mass of nitric acid $=63$
$31 g$ of phosphorous will consume $=6 g$ of nitric acid.
View full question & answer→Question 162 Marks
Calculate the percentage of boron $(B)$ in borax $(Na_2B_4O_7.10H_2O). [H = 1, B = 11, O = 16, Na = 23],$
answer correct to $1$ decimal place.
AnswerMolecular mass of borax $Na_2B_4O_7.10H_2O = 382$
$382$ parts by weight of borax contain $44$ parts by weight of boron
So$,100$ parts will contain $= 44 \times 100/382 = 11.5\%$
Percentage of boron $(B)$ in borax $(Na_2B_4O_7.10H_2O) = 11.5\%.$
View full question & answer→Question 172 Marks
$(NH_4)_2Cr_2O_7\rightarrow N_2 + Cr_2O_3 +4H_2O.$
What volume of nitrogen at $\text{STP},$ will be evolved when $63g$ of ammonium dichromate is decomposed?
$(H= 1, N = 14, 0 = 16, Cr = 52)$
AnswerMolecular mass of ammonium dichromate $= 252$
Now$, 252g$ of ammonium dichromate evolves $= 22.4 L$ of nitrogen at $\text{STP}$
$63g$ of ammonium dichromate will evolve $= 22.4 \times 63 /252 = 5.6L$
So$, 63g$ of ammonium dichromate will evolve $5.6L$ of oxygen.
View full question & answer→Question 182 Marks
What is the mass of nitrogen in 1000Kg of urea [CO(NH2)2] ?
[H = 1, C= 12, N= 14, O = 16]
AnswerMolecular mass of urea is =60
60kg of urea has = 28 Kg of nitrogen
1000Kg of urea will have = 28 x 1000/60 = 466.66 or 467 Kg.
View full question & answer→Question 192 Marks
Define or explain the meaning of term 'molar volume'.
Answer"The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure". It has been noticed that one mole of any gaseous molecules occupy 22.4 L of volume at standard temperature and pressure.
View full question & answer→Question 202 Marks
Find the total percentage of oxygen in magnesium nitrate crystal $Mg(NO_3)_2.6H_2O.$
$[H = 1, N = 14, O = 16, Mg = 24]$
AnswerMolecular mass of $Mg (NO_3)_2.6H_2O$ is $= 256$
Now$, 256$ parts by weight of crystal contains $192$ parts by weight of oxygen.
So $100$ parts by weight will contain $= 192 \times 100/256 = 75\%$
Hence, total percentage of oxygen in magnesium nitrate crystal $Mg(NO_3)_2.6H_2O$ is $75\%$
View full question & answer→Question 212 Marks
What volume of hydrogen sulphide at $\text{STP}$ will burn in oxygen to yield $12.8g$ what volume of oxygen would be required for complete combustion?
AnswerFrom the equation we know that $2\ moles$ of $H2S$ burn in presence of $3\ moles$ of Oxygen so :
$44.8L$ of $H_2S$ requires $= 67.2L$ of oxygen
$4.48L$ of $H_2S$ will require $= 67.2/44.8 \times 4.48 = 6.72L$
So$, 6.72L$ of oxygen would be required for complete combustion .
View full question & answer→Question 222 Marks
What volume of hydrogen sulphide at $\text{STP}$ will burn in oxygen to yield $12.8g$ suIphurdioxide according to the equation?
$2H_2S + 3O_2\rightarrow 2H_2O + 2SO_2$
Answer$64g$ of $SO_2$ will be produced at $\text{STP}$ from $= 22.4L$ of $H_2S$
$12.8g$ of $SO_2$ will be produced at $\text{STP}$ from $= 22.4 \times 12.8/64$
$= 4.48L$
So$, 4.48L$ of hydrogen sulphide at $\text{STP}$ will burn in oxygen to yield $12.8g$ sulphur dioxide.
View full question & answer→Question 232 Marks
Determine the molecular mass of a gas if 5g of it occupy a volume of 4 L at STP.
Answer4L of a gas at STP has mass = 5g
22.4 L of a gas at STP will has molecular mass = 5 x 22.4 / 4 = 28.
So, the molecular mass of a gas will be 28.
View full question & answer→Question 242 Marks
What is the volume occupied by 16g of sulphur dioxide. gas at STP?
AnswerVolume occupied by $16 g$ of sulphur dioxide.
$64 g$ of sulphur dioxide at STP occupies volume of $=22.4 L$
$16 g$ of sulphur dioxide at STP occupies volume of $=22.4\times 16 / 64=5.6 L$
Hence, $16 g$ of sulphur dioxide will occupy a volume of $5.6 L$
View full question & answer→Question 252 Marks
What is the volume occupied by 48g of oxygen gas at STP?
AnswerVolume occupied by 48g of oxygen:
As 32g of oxygen at STP occupies volume of = 22.4 L
48 g of oxygen at STP occupies volume of = 22.4 x 48/32 = 33.6 L
Hence, 48g of sulphur dioxide will occupy a volume of 33.6L
View full question & answer→Question 262 Marks
Potassium nitrate on strong heating decomposes as under :
$2KNO_3\rightarrow 2KNO_2 + O_2$
Calculate : Weight of potassium nitrite formed.
$(K = 39, 0 = 16, N = 14)$
AnswerFrom the equation :
Molecular weight of $KNO_3 = ($Atomic mass of $K +$ Atomic mass of $N +$ Atomic mass of $O) = (39 + 14 + 16 \times 3 ) = 101$
Molecular mass of of $KNO_2 = (39 + 14 + 16 \times 2) = 85$
From the reaction :
$2\ moles$ of $KNO_3$ gives $= 2 \ moles$ of $KNO_2$
So$, 202g$ of $KNO_3$ gives $= 170g$ of $KNO_2$
View full question & answer→Question 272 Marks
Calculate the relative molecular mass of Chloroform.
(use K = 39, Cl = 35.5, O = 16, C = 12, H = 1, Na = 23, N = 14, S= 32)
AnswerThe relative molecular mass of Chloroform $\left( CH _3 Cl \right)$ is:
[atomic mass of $1 C$ atom + Atomic mass of $3 H$ atom + atomic mass of $1 Cl$ ] $12+3 \times 1+35.5=50.5$
View full question & answer→Question 282 Marks
Calculate the relative molecular mass of Ammonium sulphate.
$($use $K = 39, Cl = 35.5, O = 16, C = 12, H = 1, Na = 23, N = 14, S= 32)$
AnswerThe relative molecular mass of Ammonium sulphate $(NH_4)_2SO_4$ is :
$[$atomic mass of $2 N$ atom $+$ Atomic mass of $8 H$ atom $+$ atomic mass of $1 S\ +$ atomic mass of $4 O$ atom$]$
$14 \times 2 + 8 \times 1 + 32 + 16 \times 4 = 132.$
View full question & answer→Question 292 Marks
Calculate the relative molecular mass of Sodium acetate
$($use $K = 39, Cl = 35.5, O = 16, C = 12, H = 1, Na = 23, N = 14, S= 32)$
AnswerThe relative molecular mass of Sodium acetate $(CH_3COONa )$ is :
$[$atomic mass of $2 C$ atom $+$ Atomic mass of $3 H$ atom $+$ atomic mass of $2 O$ atoms $+$ atomic mass of $1\ Na$ atom$]$
$12 \times 2 + 3 \times 1 + 16 \times 2 + 23 = 82$
View full question & answer→Question 302 Marks
Explain the term : Gram mole
AnswerGram mole: "A sample of substance with its mass equal to its gram molecular mass is called one gram molecule of this substance or one gram mole".
For example: Gram molecular mass of oxygen is 32g. So One gram mole of oxygen is 32g.
View full question & answer→Question 312 Marks
Explain the term : Gram atom
AnswerGram atom: "The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element".
For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram atom of hydrogen.
View full question & answer→Question 322 Marks
Define the term : Avogadro's number
AnswerAvogadro's number : Avogadro's number is defined as the number of atoms present in $12g$ of $C^{12}$ isotope i.e. $6.023 \times 10^{23}$ atoms.
- It is number of elementary units i .e. atoms, ions or molecules present in one $mole$ of a substance.
- It is denoted by $N_{A.}$
View full question & answer→Question 332 Marks
Define the term : Relative atomic mass.
AnswerRelative atomic mass : " The relative atomic mass or atomic weight of an element is the number of times one atom of the element is heavier than $1/12$ times of the mass of an atom of carbon $- 12$" .
Relative atomic mass $=$ Mass of $1$ atom of the element $1/12$ of the mass of one $C^{12}$ atom.
View full question & answer→Question 342 Marks
Define the term : Molar volume.
AnswerMolar volume : One mole of any gaseous molecules occupy $22.4\ dm^3$ at standard temperature and pressure $(\text{STP}).$ This volume is known as molar volume.
"The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure."
View full question & answer→Question 352 Marks
Define the term: Vapour density.
AnswerVapour density : It is Density of a gas, expressed as the mass of a given volume of the gas divided by the mass of an equal volume of a reference gas (such as hydrogen or air) at the same ternperature and pressure.
View full question & answer→Question 362 Marks
What is the value of molar volume of a gas at $\text{STP}$?
AnswerThe value of molar volume of a gas at $\text{STP}$ is $22.4\ dm^3($litre$)$ or $22400 \ cm^3(ml).$
Concept Insight : One mole of any gaseous molecules occupy $22.4 \ dm^3$ at standard temperature and pressure $(\text{STP})$. This volume is known as molar volume.
View full question & answer→Question 372 Marks
The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.
AnswerThe relative atomic mass of Cl atom is 35.5 a.m.u. because chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio of 3 : 1.
The average of the isotopic masses is 35 x 3 + 37 x 4 = 35.5
Concept Insight: Isotopes are atoms of same having same atomic number but different mass number.
View full question & answer→Question 382 Marks
When stating the volume of a gas, the pressure and temperature should also be given. Why?
AnswerWhen stating the volume of a gas, the pressure and temperature should also be given because the volume of a gas is highly susceptible to slight change in pressure and temperature of the gas.
View full question & answer→Question 392 Marks
Differentiate between $2H$ and $H_2$
Answer
| $2H$ |
$H_2$ |
| $2H$ corresponds to two atoms of Hydrogen element. |
$H_2$ corresponds to a hydrogen olecule which contains two atoms of hydrogen . |
View full question & answer→Question 402 Marks
What do you understand by the term mole? How many elementary units are in one mole of a substance?
Answer"A mole is defined as the amount $($mass$)$ of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i .e. $6.023 \times 10^{23}.$
Or a collect ion of $6.023 \times 10^{23}$ particles is called mole.
Number of elementary uni ts present in one mole of a substance is $6.023 \times 10^{23}.$
View full question & answer→Question 412 Marks
Prove the Following :
One mole of any gas contains the same number of molecules.
AnswerSince by definition of a mole it is defined as the amount $($mass$)$ of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i .e. $6 .023 \times 10^{23}$ so one mole of any gas contains the same number of molecules.
View full question & answer→Question 422 Marks
Give three pieces of information conveyed by the formula $H_2O.$
AnswerThree pieces of information conveyed by the formula $H_2O$ is that $:$
It shows that there are $2$ hydrogen atoms and 1oxygen atoms present in $H_2O.$ The hydrogen and oxygen atoms are present in simplest whole number ratio of $2:1.$ It represents one molecule of compound water.
View full question & answer→Question 432 Marks
Prove the Following :
2 X V.D. = Molecular mass.
AnswerThe molecular mass of the given compound is determined experimentally by vapour density method also, in which the vapour density of the compound is determined. Vapour density is related to molecular mass as :
Molecular mass = 2 x vapour density.
View full question & answer→Question 442 Marks
The equation $4NH_35O_{2 }\rightarrow 4NO + 6H_2O$ represents the catalytic oxidation of ammonia. If $100 \ cm^3$ of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.
Answer$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$
$4$ vol $: 5$ vol $4$ vol $:$ Nil
If $4$ volumes of $NH_3$ requires $=$ $\frac{5}{4} \times 100$
$= 125 \ cm^3$ oxygen.
View full question & answer→Question 452 Marks
Complete the following calculations. Show working for complete credit :
If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.
Answer6 litres of hydrogen and 4 litres of chlorine when mixed result in the formation of 8 litres of HCl gas.
When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leaving behind only 2 litres of hydrogen gas.Therefore, the volume of the residual gas will be 2 litres.
View full question & answer→Question 462 Marks
Complete the following calculations. Show working for complete credit :
Calculate the mass of calcium that will contain the same number of atoms as are present in $3.2 \ gm$ of Sulphur. $[$Atomic masses $: S = 32, Ca = 40]$
Answer$3.2g$ of $S$ has number of atoms $= 6.023 \times 10^{23} \times 3.2 /32$
$= 0.6023 \times 10^{23}$
So$, 0.6023 \times 10^{23}$ atoms of $Ca$ has mass
$= 40 \times 0.6023 \times 10^{23}/6.023 \times 10^{23} = 4g$
View full question & answer→Question 472 Marks
AnswerAvogadro's law : Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
View full question & answer→Question 482 Marks
Solve the following :
A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?
AnswerAccording to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = X.
View full question & answer→Question 492 Marks
State: Gay-Lussac's law of combining volumes.
AnswerWhen gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure.
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