Question 12 Marks
560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:
$2CO + O_2 \rightarrow 2CO_2$
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Answer$2CO + O_2 \rightarrow 2CO_2$
$2 V 1 V 2 V$
1 mole of $O _2$ has volume $=22400 ml$
The volume of oxygen used by $2 \times 22400 ml CO =22400 ml$
So, Vol. of $O _2$ used by $560 ml CO =22400 \times 560 /(2 \times 22400)$
$=280 ml$
So, the Volume of $CO _2$ formed is 560 ml .
View full question & answer→Question 22 Marks
Calculate the percentage of sodium in sodium aluminium fluoride $(Na_3AlF_6)$ correct to the nearest whole number.
$(F = 19; Na =23; Al = 27)$
AnswerMolecular mass of $Na_3AlF_6= 210$
So, Percentage of Na $= 3 x 23 x 100/210 = 32.85\%$
View full question & answer→Question 32 Marks
An Experiment showed that in a lead chloride solution, $6.21$ g of lead is combined with $4.26$ g of chlorine. What is the empirical formula of this chlorine? $( Pb =207 ; Cl =35.5)$
AnswerGram atoms of $Pb = 6.21/207 = 0.03 = 1$
Gram atoms of$ Cl = 4.26/35.5 = 0.12 = 4$
So, the empirical formula $= PbCl_4$
View full question & answer→Question 42 Marks
When heated, potassium permanganate decomposes according to the following equation :
$2 KMnO _4 \longrightarrow K _2 MnO _4+ MnO _2+ O _2$
solid residue
(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of $1.32 g$. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of $0.0825 g$, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158 . What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of $15.8 g$ of potassium permanganate? (Molar volume at room temperature is 24 litres)
Answer(a) $1$ litre of oxygen has mass $=1.32 g$
So, 24 litres (molar vol. at room temp.) will have mass $=1.32 \times 24$ $=31.6$ or 32 g
(b) $2 KMnO _4 \rightarrow K_2 MnO _4+ MnO _2+ O _2$
316 g of $KMnO _4$ gives oxygen $=24$ litres
So, $15.8$ g of $KMnO _4$ will give $=24 \times 316 / 15.8=1.2$ litres
View full question & answer→Question 52 Marks
The percentage composition of sodium phosphate as determined by analysis is $42.1 \%$ sodium, $18.9 \%$ phosphorus and $39 \%$ oxygen. Find the empirical formula of the compound?
Answersimple ratio of $Na = 42.1/23 = 1.83 = 3$
simple ratio of $P = 18.9/31 = 0.609 = 1$
simple ratio of $O = 39/16 = 2.43 = 4$
So, the empirical formula is $Na_3PO_4$
View full question & answer→Question 62 Marks
What is the mass of $56 cm^3$ of carbon monoxide at $S.T.P?$
Answer$22400 cm^3$ volume have mass $=28 g$ of CO (molar mass)
So, $56 cm^3$ volume will have mass $=28 \times 56 / 22400=0.07 g$
View full question & answer→Question 72 Marks
What is the volume at $S.T.P$. of $7.1$ g of chlorine?
AnswerThe volume of 71 g of $Cl _2$ at STP $=22.4$ litres
Volume of 7.1 g chlorine $=22.4 \times 7.1 / 71=2.24$ litre
View full question & answer→Question 82 Marks
A compound $X$ consists of $4.8 \%$ carbon and $95.2 \%$ bromine by mass.
If the vapour density of the compound is $252$, what is the molecular formula of the compound?
AnswerEmpirical formula mass $= 12 + 3 × 80 = 252 g$
molecular formula mass $= 2 \times 252(V.D) = 504 g$
$n = 504/252 = 2$
so, molecular formula $= C_2Br_6$
View full question & answer→Question 92 Marks
A compound $X$ consists of $4.8 \%$ carbon and $95.2 \%$ bromine by mass.
Determine the empirical formula of this compound working correctly to one decimal place ( $C =12 ; Br =80$ )
AnswerElement % atomic mass atomic ratio simple ratio
C 4.$812 `4.8/12 = 0.4_1`$
Br$ 95.280 `95.2/80 = 1.2_3`$
So, empirical formula is $CBr_3$
View full question & answer→Question 102 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 $(M=56)$.
AnswerSimple ratio of $M = 34.5/56 = 0.616 = 1$
Simple ratio of $Cl = 65.5/35.5 = 1.845 = 3$
Empirical formula $= MCl_3$
Empirical formula mass = 162.5, Molecular mass $= 2 \times V.D = 325$
So, $n = 2$
So, molecular formula $= M_2Cl_6$
View full question & answer→Question 112 Marks
Washing soda has formula $Na _2 CO _3 \cdot 10 H _2 O$. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
Answer$Na_2CO_3.10H_2O \rightarrow Na_2CO_3 + 10H_2O$
$286 \ g 106 \ g$
So, for $57.2 g Na_2CO_3.10H_2O = 106 \times 57.2/286 = 21.2 g Na_2CO_3$
View full question & answer→Question 122 Marks
If a crop of wheat removes $20 \ kg$ of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate $Ca(NO_3)_2$ would be required to replace the nitrogen in a 10-hectare field?
AnswerIn 1 hectare of soil, $N _2$ removed $=20 kg$
So, in 10 hectare $N _2$ removed $=200 kg$
The molecular mass of $Ca \left( NO _3\right)_2=164$
Now, $28 g N N _2$ present in fertilizer $=164 g Ca \left( NO _3\right)_2$
So, 200000 g of $N _2$ is present in $=164 \times 200000 / 28$
$=1171.42 kg$
View full question & answer→Question 132 Marks
$1 $g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.
$AgNO_3(aq)+NaCl(aq) \rightarrow AgCl(s)+NaNO_3$
Calculate the precentage of NaCl in the mixture.
AnswerSince 143.5 g of AgCl is produced from $=58.5 g$ of NaCl
so, 1.435 g of AgCl is formed by $=0.585 g$ of NaCl
$\%$ of $NaCl =0.585 \times 100=58.5 \%$
View full question & answer→Question 142 Marks
A hydrate of calcium sulphate $CaSO _4 \cdot xH _2 O$ contains $21 \%$ water of crystallization. Find the value of $x$.
AnswerThe total molar mass of hydrated $CaSO _4 \cdot xH _2 O =136+18 x$
Since $21 \%$ is water of crystallization, so
$`(18"x")/(136 + 18"x") = 21/100`$
So, $x=2$ i.e. water of crystallization is 2 .
View full question & answer→Question 152 Marks
From the equation for burning of hydrogen and oxygen $2 H _2+ O _2 \longrightarrow 2 H _2 O$ (Steam)
Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.
AnswerFrom equation: $2H_2 + O_2 \rightarrow 2H_2O$
$1$ mole of Oxygen gives $= 2$ moles of steam
so, $0.5$ mole oxygen will give $= 2 \times 0.5 =$ 1mole of steam
View full question & answer→Question 162 Marks
An acid of phosphorus has the following percentage composition; Phosphorus $=38.27 \%$; hydrogen $=2.47 \%$;; oxygen $= 59.26 \%.$ Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is $162.$
Answer$\%$ composition Atomic ratio Simple ratio
$P = 38.27%$
$\Rightarrow 38.27/31 =1.23 : 1$
$H = 2.47%$
$\Rightarrow 2.47/1 = 2.47 : 2$
$O = 59.26%$
$\Rightarrow 59.26/16 = 3.70 : 3$
So, empirical formula is $PH_2O_3 or H_2PO_3$
Empirical formula mass $= 31+ 2 × 1 + 3 × 16 = 81$
The molecular formula is $= H_4P_2O_6$, because $n = 162/81=2$
View full question & answer→Question 172 Marks
When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :
$CO_2 + C \rightarrow 2CO$
What volume of carbon monoxide at $S.T.P$. can be obtained from $3 g$ of carbon?
Answer$CO_2 + C \rightarrow 2CO$
$1V 1V 2V$
$12 g$ of C gives $=44.8$ litre volume of CO
So,$ 3 g$ of C will give $=11.2$ litre of CO
View full question & answer→Question 182 Marks
The reaction between the red lead and hydrochloric acid is given below:
$Pb_3O_4+ 8HCl \rightarrow 3PbCl_2+ 4H_2O + Cl_2$
Calculate: the volume of the chlorine evolved at $S.T.P.$
Answer$1 V Pb_3O_4$produces $1 V Cl_2$
$685g$ of $Pb_3O_4$ has volume $= 22.4$ litres = volume of $Cl_2$ produced
So, $6.85 Pb_3O_4$ will produce $= 6.85 22.4/685 = 0.224$ litres of $Cl_2$
View full question & answer→Question 192 Marks
The reaction between the red lead and hydrochloric acid is given below:
$Pb_3 O_4+8 HCl \rightarrow 3 PbCl_2+4 H_2 O+Cl_2$
Calculate: the mass of the chlorine
Answer$685\ g$ of $Pb _3 O _4$ gives $=71 g$ of $Cl _2$
Hence, $6.85\ g$ of $Pb _3 O _4$ will give $=6.85 \times 71 / 685=0.71\ g Cl _2$
View full question & answer→Question 202 Marks
The reaction between the red lead and hydrochloric acid is given below:
$Pb_3 O_4+8 HCl \rightarrow 3 PbCl_2+4 H_2 O+Cl_2$
Calculate: the mass of lead chloride formed by the action of the $6.85\ g$ of red lead
AnswerMolecular mass of $Pb _3 O _4=3207.2+416=685\ g$
$685\ g$ of $Pb _3 O _4$ gives $=834 g$ of $PbCl _2$
Hence, $6.85\ g$ of $Pb _3 O _4$ will give $=6.85834 / 685=8.34 g$
View full question & answer→Question 212 Marks
The reaction between $15\ g$ of marble and nitric acid is given by the following equation:
$CaCO_3+2 HNO_3 Ca\left(NO_3\right)_2+H_2 O+CO_2$
Calculate: the volume of carbon dioxide evolved at S.T.P.
Answer$1\ V$ of $CaCO _3$ produces $1\ V$ of $CO _2$
$100\ g$ of $CaCO _3$ has volume $=22.4$ litres
So, $15\ g$ will have volume $=22.415 / 100=3.36$ litres $CO _2$
View full question & answer→Question 222 Marks
The reaction between $15\ g$ of marble and nitric acid is given by the following equation:
$CaCO_3+2 HNO_3 Ca\left(NO_3\right)_2+H_2 O+CO_2$
Calculate: the mass of anhydrous calcium nitrate formed
Answer$100\ g$ of $CaCO _3$ produces $=164 g$ of $Ca \left( NO _3\right)_2$
So, $15\ g\ CaCO _3$ will produce $=164 15 / 100=24.6 g Ca \left( NO _3\right)_2$
View full question & answer→Question 232 Marks
$MnO2 + 4HCl \rightarrow MnCl2 + 2H2O +Cl2$
$0.02$ moles of pure MnO2is heated strongly with conc. $HCl$. Calculate: moles of acid required
Answer$MnO_2+4 HCl \rightarrow MnCl_2+2 H_2 O+Cl_2$
$1 V 4 V 1 V 1 V$
$1{molele MnO_2 \text { requires } HCl=4 \text { mole }}_2^l$
So, $0.02$ molele ${ MnO _2}$ will require $=40.02=0.08 mole$
View full question & answer→Question 242 Marks
$MnO_2+4 HCl \rightarrow MnCl_2+2 H_2 O+Cl_2$
$0.02$ moles of pure $MnO _2$ is heated strongly with conc. $HCl$ . Calculate volume of chlorine gas formed at $S.T.P.$
Answer$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
$1$ mole of chlorine gas has volume $= 22.4$ litres
So, $0.02$ mole will have volume $= 22.4 0.02 = 0.448$ litre
View full question & answer→Question 252 Marks
$$MnO_2+4 HCl \rightarrow MnCl_2+2 H_2 O+Cl_2$
$0.02$ moles of pure $MnO _2$ is heated strongly with conc. $HCl$ . Calculate: the mass of chlorine gas formed
Answer$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
$1$ mole of $Cl_2$ weighs $= 35.5 g$
So, $0.02$ mole will weigh $= 71 0.02 = 1.42 g$ of $Cl_2$
View full question & answer→Question 262 Marks
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$0.02$ moles of pure $MnO_2$is heated strongly with conc. $HCl$. Calculate: moles of chlorine gas formed
Answer$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
$0.02 mole_{ MnO }^2$ will form $=0.02 mole$ of $Cl _2$
View full question & answer→Question 272 Marks
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$0.02$ moles of pure $MnO_2$is heated strongly with conc. $HCl.$ Calculate: the mass of salt formed
Answer$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
$1$ mole $MnCl _2$ weighs $=126 g$ (mol mass)
So, $0.02 mole MnCl _2$ will weigh $=1260.02 g=2.52 g$
View full question & answer→Question 282 Marks
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$0.02$ moles of pure $MnO_2$ is heated strongly with conc. $HCl$. Calculate: moles of salt formed
Answer$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
1 mole $MnO_2$ gives $= 1$ mole of $MnCl_2$
So, $0.02$ mole $MnO_2$will give = $0.02$ mole of $MnCl_2$
View full question & answer→Question 292 Marks
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O +Cl_2$
$0.02$ moles of pure $MnO_2$ is heated strongly with conc. $HCl$. Calculate:
mass of $MnO_2$ used
Answer$MnO_2 + 4HCl MnCl_2 + 2H_2O +Cl_2$
$1 V4 V1 V1 V$
$1 $mole of $MnO_2$ weighs $= 87 g$ (mol. Mass)
So, $0.02$ mole will weigh $= 87 0.02 = 1.74 g MnO_2$
View full question & answer→Question 302 Marks
Calculate the percent composition of Potassium chlorate $KClO_3.$
AnswerMolecular mass of $KClO _3=122.5 g$
$\%$ of $K =39 / 122.5=31.8 \%$
\% of $Cl =35.5 / 122.5=28.98 \%$
$\%$ of $O=316 / 122.5=39.18 \%$
View full question & answer→Question 312 Marks
Explain the terms empirical formula and molecular formula.
AnswerThe empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.
View full question & answer→Question 322 Marks
A hydride of nitrogen contains $87.5\%$ percent by mass of nitrogen. Determine the empirical formula of this compound.
AnswerAtomic ratio of $N =87.5 / 14=6.25$
Atomic ratio of $H =12.5 / 1=12.5$
This gives us the simplest ratio as 1:2
So, the molecular formula is $NH _2$
View full question & answer→Question 332 Marks
A compound with empirical formula $AB$ has vapour density $3$ times its empirical formula weight. Find the molecular formula.
AnswerSince molecular mass is $2$ times the vapour density, so Mol. Mass $= 2 V.D$
Empirical formula weight $= V.D/3$
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is $A_6B_6$
View full question & answer→Question 342 Marks
Determine the formula of the organic compound if its molecule contains $12$ atoms of carbon. The percentage compositions of hydrogen and oxygen are $6.48$ and $51.42$ respectively.
AnswerElement $\%$ At. mass Atomic ratio Simple ratio
$C 42.1 12 3.5 1$
$H 6.48 1 6.48 2$
$O 51.42 16 3.2 1$
The empirical formula is $CH_2O$
Since the compound has $12$ atoms of carbon, so the formula is
$C_{12} H_{24} O_{12.}$
View full question & answer→Question 352 Marks
Urea is a very important nitrogenous fertilizer. Its formula is $CON _2 H _4$. Calculate the percentage of nitrogen in urea. $( C =12, O =16, N=14$ and $H =1$ ).
AnswerMolar mass of urea; $CON_2H_4= 60 g$
So, $\%$ of Nitrogen $= 28 \times 100/60 = 46.66\%$
View full question & answer→Question 362 Marks
In a compound of magnesium $( Mg =24)$ and nitrogen $( N =14), 18 g$ of magnesium combines with 7 g of nitrogen. Deduce the simplest formula by answering the following questions.
(a) How many gram-atoms of magnesium are equal to $18 g ?$
(b) How many gram-atoms of nitrogen are equal to $7$ g of nitrogen?
(c) Calculate the simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.
Answer(a) G atoms of magnesium $=18 / 24=0.75$ or g - atom of $Mg$
(b) G atoms of nitrogen $=7 / 14=0.5$ or $1 / 2 g$-atoms of $N$
(c) Ratio of gram-atoms of N and $Mg =1: 1.5$ or $2: 3$
So, the formula is $Mg_3N_2$
View full question & answer→Question 372 Marks
A hydrocarbon contains $4.8g$ of carbon per gram of hydrogen. Calculate Find molecular formula, if its vapour density is $29.$
AnswerEmpirical formula mass $= 29$
Molecular mass $= V.D \times 2 = 29 \times 2 = 58$
So, molecular formula $= C_4H_{10}$
View full question & answer→Question 382 Marks
A hydrocarbon contains $4.8g$ of carbon per gram of hydrogen. Calculate find the empirical formula
AnswerElement Given mass At. mass Gram atom Ratio
$C 4.8 12 0.4 1 2$
$H 1 1 1 2.5 5$
So, the empirical formula $= C_2H_5$
View full question & answer→Question 392 Marks
Calculate the number of
Hydrogen atoms in $0.1$ mole of $H _2 SO _4$.
Answer1 mole of $H _2 SO _4$ contains $=2 \times 6.023 \times 10^{23}$
So, 0.1 mole of $H _2 SO _4$ contains $=2 \times 6.023 \times 10^{23} \times 0.1$
$=1.2 \times 10^{23}$ atoms of hydrogen
View full question & answer→Question 402 Marks
Which of the following would weigh most?
(a) $1$ mole of $H_2O$
(b) $1$ mole of $CO_2$
(c) $1$ mole of $NH_3$
(d) $1$ mole of $CO$
AnswerMolecular mass of $H _2 O$ is $18, CO _2$ is $44, NH _3$ is 17 and $CO$ is $28$
So, the weight of 1 mole of $CO _2$ is more than the other three.
View full question & answer→Question 412 Marks
Which of the following weighs the least?
$3 x10^{25}$ atoms of carbon
Answer$6.023 x10^{23}$ atoms of C weigh $= 12 g$
$So, 3 x10^{25}$ atoms will weigh $= `(12 xx 3 xx 10^25)/(6.023 xx 10^23) = 597.7`g$
View full question & answer→Question 422 Marks
A student puts his signature with a graphite pencil. If the mass of carbon in the signature is $10^{-12} g$, calculate the number of carbon atoms in the signature.
AnswerNo. of atoms in 12 g $C = 6.023 x10^{23}$
So, no. of carbon atoms in $10^{-12} g = 10^{-12} x 6.023 x10^{23}/12$
$= 5.019 x 10^{10}$atoms
View full question & answer→Question 432 Marks
Calculate:
The gram molecular mass of chlorine if $308cm^3$ of it at $STP$ weighs $0.979 g$
Answer$308 cm^3$ of chlorine weighs $=0.979 g$
So, $22400 cm^3$ will weigh $=$ gram molecular mass
$=0.979 \times 22400 / 308=71.2 g$
View full question & answer→Question 442 Marks
The molecular formula for elemental sulphur is $S_8$.In sample of $5.12 g$ of sulphur
How many molecules and atoms are present?
AnswerNo. of molecules $= 0.02 x 6.023 x 10^{23} = 1.2 x 10^{22}$ molecules
No. of atoms in $1$ molecule of $S = 8$
So, no. of atoms in $1.2 x 10^{22}$ molecules $= 1.2 x 10^{22}x 8$
$= 9.635x 10^{22}$ molecules
View full question & answer→Question 452 Marks
Determine the number of molecules in a drop of water which weighs $0.09 g .$
Answer18 g of water has number of molecules $=6.023 \times 10^{23}$
So, 0.09 g of water will have no. of molecules $=6.023 \times 10^{23} \times 0.09 / 18=3.01 \times 10^{21}$ molecules
View full question & answer→Question 462 Marks
What is the mass of $56 cm^3$ of carbon monoxide at $STP?$
$(C=12 ,O=16)$
Answer$22400 cm^3$ of CO has mass $=28 g$
So, $56 cm^3$ will have mass $=56 \times 28 / 22400=0.07 g$
View full question & answer→Question 472 Marks
Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. [S = 32; O = 16]
Answer64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres
View full question & answer→Question 482 Marks
Calculate the mass of nitrogen supplied to soil by $5 \ kg$ of urea $[CO(NH_2)_2] [O = 16; N = 14; C = 12 ; H = 1 ]$
Answer$60$ g urea has mass of nitrogen $(N_2) = 28 g$
So, $5000$ g urea will have mass $= 28 x 5000/60 = 2.33 kg$
View full question & answer→Question 492 Marks
Calculate the number of atoms Which is given below :
$52$ g of He
Answer$4$ g of He has atoms $= 6.023 x10^{23}$
So, $52$ g will have $= `(6.023 xx 10^23 xx 52)/4 = 7.828 xx 10^24`$ atoms
View full question & answer→Question 502 Marks
Calculate the number of oxygen atoms in 0.10 mole of $Na _2 CO _3 \cdot 10 H _2 O$.
Answer1molecule of $Na_2CO_3.10H_2O$ contains oxygen atoms $= 13$
So,$ 6.023 x 10^{23}$ molecules (1mole) has atoms $=13 x 6.023 x 10^{23}$
So, $0.1$ mole will have atoms $= 0.1 x 13 x 6.023 x 10^{23} =7.8 x 10^{23}$
View full question & answer→