Question 11 Mark
Calculate the number of moles in 7g of nitrogen.
Answer28g of nitrogen = 1mole
So, 7g of nitrogen = 1/28 × 7= 0.25 moless
View full question & answer→Question 21 Mark
Ordinary chlorine gas has two isotopes ${ }^{35}{ }_{17} Cl$ and ${ }^{37}{ }_{17} Cl$ in the ratio of $3: 1$. Calculate the relative atomic mass of chlorine.
AnswerThe relative atomic mass of $Cl =(35 \times 3+1 \times 37) / 4=35.5 amu$
View full question & answer→Question 31 Mark
A gas cylinder filled with hydrogen holds $5 g$ of the gas. The same cylinder holds $85 g$ of gas $X$ under the same temperature and pressure. Calculate :
The molecular weight of gas $X$.
AnswerMolecular mass = 17(V.D) x 2 = 34g
View full question & answer→Question 41 Mark
A gas cylinder filled with hydrogen holds 5g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate :
Vapour density of gas X.
AnswerV.D = `" mass of gas at STP"/("mass of the equal volume of" "H"_2) = 85/5 = 17`
View full question & answer→Question 51 Mark
Calculate the percentage of phosphorus in Calcium phosphate $Ca _3\left( PO _4\right)_2$
AnswerMolecular mass of $Ca _3\left( PO _4\right)_2=310$
$\%$ of $P=231$ $100/310 = 20$%
View full question & answer→Question 61 Mark
Calculate the percentage of phosphorus in Calcium hydrogen phosphate $Ca(H_2PO_4)_2$
AnswerMolecular mass of $Ca \left( H _2 PO _4\right)_2=234$
So, % of $P=231$ $100/234 = 26.5$ %
View full question & answer→Question 71 Mark
Give the empirical formula of: $CH_3COOH$
AnswerThe empirical formula of $CH _3 COOH$ is: $C H _2 O$.
View full question & answer→Question 81 Mark
Give the empirical formula of : $C_2H_2$
AnswerThe empirical formula of $C _2 H _2$ is: $C H$
View full question & answer→Question 91 Mark
Give the empirical formula of: $C_6H_8O_3$
View full question & answer→Question 101 Mark
Give the empirical formula of : $C_6H_6$
AnswerThe empirical formula of $C _6 H _8$ is: $CH$
View full question & answer→Question 111 Mark
A compound with empirical formula $AB _2$ has the vapour density equal to its empirical formula weight. Find its molecular formula.
AnswerNow since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have $n=2$ so, the molecular formula is $A _2 B_4$.
View full question & answer→Question 121 Mark
A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate the g atom of each
AnswerThe g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
View full question & answer→Question 131 Mark
How many grams of
$CO _2$ is present in 0.1 mole of it?
Answer$0.1$ mole of $CO _2$ has mass $=0.1 \times 44=4.4 g$
View full question & answer→Question 141 Mark
How many grams of
$H _2 O$ are present in $0.2$ mole of it?
Answer$0.2$ mole of $H _2 O$ has mass $=0.2 \times 18=3.6 g$
View full question & answer→Question 151 Mark
How many grams of
HCl are present in 0.1 mole of it?
Answer0.1 mole of HCl has mass = 0.1 x 36.5(mass of 1 mole)
= 3.65 g
View full question & answer→Question 161 Mark
How many grams of
Al are present in 0.2 mole of it?
Answer1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
View full question & answer→Question 171 Mark
Calculate the number of
Molecules in one $Kg$ of calcium chloride.
Answer$111 g CaCl _2$ contains $=6.023 \times 10^{23}$ molecules
So, 1000 g contains $=5.42 \times 10^{24}$ molecules
View full question & answer→Question 181 Mark
Calculate the number of
Particles in $0.1$ mole of any substance.
AnswerNo. of particles in s1 mole $=6.023 \times 10^{23}$
So, particles in $0.1 mole=6.023 \times 10^{23} \times 0.1=6.023 \times 10^{22}$
View full question & answer→Question 191 Mark
Find the Number of molecules in $3.2\ g$ of $SO_{2.}$
AnswerNo. of molecules in $3.2\ g$ of $SO _2=6.023 \times 10^{23} \times 3.2 / 64$
$=3.023 \times 10^{22}$
View full question & answer→Question 201 Mark
Find the Weight of $0.2$ mole of $H_2$ gas
AnswerWeight of 0.2 mole $H _2$ gas $=2$ (Mol. Mass) $\times 0.2=0.4 g$
View full question & answer→Question 211 Mark
Find the number of moles in $10\ g$ of $CaCO_3$
AnswerNo. of moles in 10 g of $CaCO _3=10 / 100\left(mol\right.$. Mass $\left.CaCO _3\right)$
$=0.1 mole$
View full question & answer→Question 221 Mark
Find the number of molecules in $1.8\ g$ of $H_2O$
AnswerNo. of molecules in $1.8 g H _2 O =6.023 \times 10^{23} \times 1.8 / 18$ $=6.023 \times 10^{22}$
View full question & answer→Question 231 Mark
Find the weight of $0.5$ mole of $O_2$
AnswerWeight of $0.5$ mole of $O_2$ is = $32$(mol. Mass of $O_2$) $\times$ $0.5=16\ g$
View full question & answer→Question 241 Mark
Find the number of molecules in 73 g of HCl
AnswerNo. of molecules in $73 g HCl =6.023 \times 10^{23} \times 73 / 36.5$ (molmass of $HCl$ )
$
=12.04 \times 10^{23}
$
View full question & answer→Question 251 Mark
Calculate the relative molecular mass of :
$
\left( NH _4\right)_2 Cr _2 O _7
$
Answer(2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252
View full question & answer→Question 261 Mark
Calculate the relative molecular mass of :
$CHCl_3$
Answer$( C ) 12+( H ) 1+(3 Cl ) 3 \times 35.5=119.5$
View full question & answer→Question 271 Mark
Calculate the relative molecular mass of :
$
CH _3 COONa
$
Answer(C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
View full question & answer→Question 281 Mark
Calculate the relative molecular mass of :
$
\left( NH _4\right)_2 SO _4
$
Answer(2N)28 + (8H)8 + (S)32 + (4O)64 = 132
View full question & answer→Question 291 Mark
Calculate the relative molecular mass of :
$CuSO_4. 5H_2O$
Answer$( Cu ) 63.5+( S ) 32+(4 O ) 64+\left(5 H _2 O \right) 5 \times 18=249.5$
View full question & answer→Question 301 Mark
Calculate the relative molecular mass of :
Potassium chlorate
Answer$KClO _3=( K ) 39+( Cl ) 35.5+(3 O ) 48=122.5$
View full question & answer→Question 311 Mark
Calculate the relative molecular mass of :
Ammonium chloroplatinate $(NH_4)_2 PtCl_6$
Answer(2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
View full question & answer→Question 321 Mark
Correct the statement, if required
Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.
AnswerUnder similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.
View full question & answer→Question 331 Mark
Correct the statement, if required
The relative atomic mass of an element is the number of times one molecule of an element is heavier than $1 / 12$ the mass of an atom of $C ^{12}$.
AnswerThe relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of `"C"^12`.
View full question & answer→Question 341 Mark
Correct the statement, if required
Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.
AnswerUnder similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
View full question & answer→Question 351 Mark
Correct the statement, if required
One mole of chlorine contains $6.023 \times 10^{10}$ atoms of chlorine.
AnswerOne mole of chlorine contains $6.023 x 10^{23}$ atoms of chlorine.
View full question & answer→Question 361 Mark
Which of the following weighs the least?
7g silver
Answer$7 g$ of silver
So, 7 grams of silver weighs the least.
View full question & answer→Question 371 Mark
Which of the following weighs the least?
1 mole of sulphur
Answer1 mole of sulphur weighs = 32 g
View full question & answer→Question 381 Mark
Which of the following weighs the least?
$2 g$ atom of $N$
AnswerWeight of $1 g$ atom $N =14 g$
So, weight of $2 g$ atom of $N =28 g$
View full question & answer→Question 391 Mark
Cost of Sugar $(C_{12}H_{22} O_{11})$ is Rs $40$ per kg; calculate its cost per mole.
Answer$
1000 g \text { of sugar costs }=\text { Rs. } 40
$
So, $342 g$ (molar mass) of sugar will cost $=342 \times 40 / 1000=$ Rs. 13.68
View full question & answer→Question 401 Mark
Define or explain the term
Mole
AnswerMole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.
View full question & answer→Question 411 Mark
Define or explain the term
Gram atom
AnswerThe quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.
View full question & answer→Question 421 Mark
Define or explain the term
Avogadro's number
AnswerThe number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. $6.023 x10^{23}$ atoms.
View full question & answer→Question 431 Mark
Define or explain the term
Relative molecular mass
AnswerThe relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon -12.
View full question & answer→Question 441 Mark
Calculate :
The mass of oxygen in $2.2$ litres of $CO_2$ at STP.
AnswerMass of oxygen in $22.4$ litres $=32 g$ (molar mass)
So, mass of oxygen in $2.2$ litres $=2.2 \times 32 / 22.4=3.14 g$
View full question & answer→Question 451 Mark
Define or explain the term
Relative atomic mass
AnswerThe relative atomic mass 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.
View full question & answer→Question 461 Mark
Define or explain the term
Molar volume
AnswerMolar volume is the volume occupied by one mole of the gas at STP.
It is equal to $22.4 dm^3.$
View full question & answer→Question 471 Mark
The molecular formula for elemental sulphur is $S_8 . I n$ sample of $5.12 g$ of sulphur
How many moles of sulphur are present?
AnswerNo. of moles in $256 g S _8=1$ mole
So, no. of moles in $5.12 g =5.12 / 256=0.02$ moles
View full question & answer→Question 481 Mark
Define or explain the term
Vapour density
AnswerThe vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.
View full question & answer→Question 491 Mark
The atomic mass of Chlorine is 35.5. What is its vapour density?
AnswerVapour density of Chlorine atom is 35.5
View full question & answer→Question 501 Mark
What do you understand by the statement that 'vapour density of carbon dioxide is 22'?
AnswerVapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
View full question & answer→