Q-20 SUMMIT
QUESTIONS
© Ransho Revision
Designed with hard work with your mentor Sankalp Jauhari
Chapter: Mole Concept
,Q-20 Summit Questions
Q1. An element Z has two isotopes, 10 Z and 11 Z. Its chloride (ZCl) is gaseous
at room temperature. The measured vapour density of ZCl is 40.5. Find the
percentage abundance of each isotope of Z.
(Use: Vapour density = Molecular mass/2).
Easy language
Idea seedha hai: VD se molar mass nikalta hai (M = 2 × VD). Phir chloride se
chlorine ka hissa minus karke Z ka average atomic mass milta hai. Do isotopes 10 Z
aur 11 Z ke mixture ke liye weighted average: AZ = 10x + 11(1 − x). Yahin se x (i.e.,
10
Z ka fraction) nikal lo.
Best approach to solve question
• Compute M (ZCl) = 2 × 40.5 = 81.0 u.
• Deduce how many chlorines are present by requiring the residual mass for Z to lie
between 10 and 11 u.
• Obtain AZ (average atomic mass of Z) from M − n × ACl .
• Let x be the fractional abundance of 10 Z (1 − x for 11 Z); solve AZ = 10x + 11(1 − x).
C O VA L E N T
Step by step solution
Step 1: Find the molar mass from vapour density
M (ZCl) = 2 × 40.5 = 81.0 u.
Step 2: Infer the chlorine count
We need M − n × ACl ∈ (10, 11) u. Trying n = 2: 81.0 − 2 × 35.5 = 10.0 u ∈ (10, 11) ⇒
n = 2. (So the gaseous chloride behaves effectively as ZCl2 under these data.)
Step 3: Average atomic mass of Z
AZ = 81.0 − 2 × 35.5 = 10.0 u.
Step 4: Isotopic abundances
Let x = fraction of 10
Z. Then
AZ = 10x + 11(1 − x) = 11 − x.
With AZ = 10.0: 11 − x = 10.0 ⇒ x = 1.00.
10 11
Z : 100% Z : 0%
Exam note: If ACl = 35.45 u is used, AZ = 81 − 2(35.45) = 10.10 u ⇒ x = 0.90, giving
≈ 10 Z : 90%, 11 Z : 10%.
Mole concept Page 1
, Q-20 Summit Questions
Q2. Two elements X and Y form compounds X2 Y and XY2 . The average
atomic mass of X is 24 u, and of Y is 35.5 u. If 0.5 mole of X2 Y and 0.5 mole
of XY2 are mixed without reaction, calculate the ratio of the total mass of X
atoms to the total mass of Y atoms in the mixture.
Easy language
Seedha logic: har compound se X aur Y ke moles count karo, phir total moles nikaal
ke atomic mass se multiply karke mass ratio nikaal do. Mass ratio ∝ (moles) ×
(atomic mass).
Best approach to solve question
• From 0.5 mol X2 Y : nX = 2 × 0.5, nY = 1 × 0.5.
• From 0.5 mol XY2 : nX = 1 × 0.5, nY = 2 × 0.5.
• Sum moles of each element; multiply by atomic masses 24 and 35.5 to get total
masses.
• Form and simplify the ratio mX : mY .
C O VA L E N T
Step by step solution
Step 1: Mole contributions from each compound
From 0.5 mol X2 Y : nX = 1.0 mol, nY = 0.5 mol.
From 0.5 mol XY2 : nX = 0.5 mol, nY = 1.0 mol.
Step 2: Total moles of atoms in the mixture
(tot) (tot)
nX = 1.0 + 0.5 = 1.5 mol, nY = 0.5 + 1.0 = 1.5 mol.
Step 3: Total masses and the desired ratio
(tot) (tot)
m X = nX × 24 = 1.5 × 24, mY = n Y × 35.5 = 1.5 × 35.5.
Hence
mX : mY = (1.5 × 24) : (1.5 × 35.5) = 24 : 35.5 = 48 : 71 .
Q3. Chlorine exists as two isotopes, 35 Cl (mass = 34.97 u, abundance = 75.5%)
and 37 Cl (mass = 36.97 u, abundance = 24.5%). A compound XCl4 is formed.
If the experimentally measured molecular mass of the compound is 153.8 u,
identify the element X.
(Take the average atomic mass of Cl correctly into account.)
Page 2
© RANSHO
QUESTIONS
© Ransho Revision
Designed with hard work with your mentor Sankalp Jauhari
Chapter: Mole Concept
,Q-20 Summit Questions
Q1. An element Z has two isotopes, 10 Z and 11 Z. Its chloride (ZCl) is gaseous
at room temperature. The measured vapour density of ZCl is 40.5. Find the
percentage abundance of each isotope of Z.
(Use: Vapour density = Molecular mass/2).
Easy language
Idea seedha hai: VD se molar mass nikalta hai (M = 2 × VD). Phir chloride se
chlorine ka hissa minus karke Z ka average atomic mass milta hai. Do isotopes 10 Z
aur 11 Z ke mixture ke liye weighted average: AZ = 10x + 11(1 − x). Yahin se x (i.e.,
10
Z ka fraction) nikal lo.
Best approach to solve question
• Compute M (ZCl) = 2 × 40.5 = 81.0 u.
• Deduce how many chlorines are present by requiring the residual mass for Z to lie
between 10 and 11 u.
• Obtain AZ (average atomic mass of Z) from M − n × ACl .
• Let x be the fractional abundance of 10 Z (1 − x for 11 Z); solve AZ = 10x + 11(1 − x).
C O VA L E N T
Step by step solution
Step 1: Find the molar mass from vapour density
M (ZCl) = 2 × 40.5 = 81.0 u.
Step 2: Infer the chlorine count
We need M − n × ACl ∈ (10, 11) u. Trying n = 2: 81.0 − 2 × 35.5 = 10.0 u ∈ (10, 11) ⇒
n = 2. (So the gaseous chloride behaves effectively as ZCl2 under these data.)
Step 3: Average atomic mass of Z
AZ = 81.0 − 2 × 35.5 = 10.0 u.
Step 4: Isotopic abundances
Let x = fraction of 10
Z. Then
AZ = 10x + 11(1 − x) = 11 − x.
With AZ = 10.0: 11 − x = 10.0 ⇒ x = 1.00.
10 11
Z : 100% Z : 0%
Exam note: If ACl = 35.45 u is used, AZ = 81 − 2(35.45) = 10.10 u ⇒ x = 0.90, giving
≈ 10 Z : 90%, 11 Z : 10%.
Mole concept Page 1
, Q-20 Summit Questions
Q2. Two elements X and Y form compounds X2 Y and XY2 . The average
atomic mass of X is 24 u, and of Y is 35.5 u. If 0.5 mole of X2 Y and 0.5 mole
of XY2 are mixed without reaction, calculate the ratio of the total mass of X
atoms to the total mass of Y atoms in the mixture.
Easy language
Seedha logic: har compound se X aur Y ke moles count karo, phir total moles nikaal
ke atomic mass se multiply karke mass ratio nikaal do. Mass ratio ∝ (moles) ×
(atomic mass).
Best approach to solve question
• From 0.5 mol X2 Y : nX = 2 × 0.5, nY = 1 × 0.5.
• From 0.5 mol XY2 : nX = 1 × 0.5, nY = 2 × 0.5.
• Sum moles of each element; multiply by atomic masses 24 and 35.5 to get total
masses.
• Form and simplify the ratio mX : mY .
C O VA L E N T
Step by step solution
Step 1: Mole contributions from each compound
From 0.5 mol X2 Y : nX = 1.0 mol, nY = 0.5 mol.
From 0.5 mol XY2 : nX = 0.5 mol, nY = 1.0 mol.
Step 2: Total moles of atoms in the mixture
(tot) (tot)
nX = 1.0 + 0.5 = 1.5 mol, nY = 0.5 + 1.0 = 1.5 mol.
Step 3: Total masses and the desired ratio
(tot) (tot)
m X = nX × 24 = 1.5 × 24, mY = n Y × 35.5 = 1.5 × 35.5.
Hence
mX : mY = (1.5 × 24) : (1.5 × 35.5) = 24 : 35.5 = 48 : 71 .
Q3. Chlorine exists as two isotopes, 35 Cl (mass = 34.97 u, abundance = 75.5%)
and 37 Cl (mass = 36.97 u, abundance = 24.5%). A compound XCl4 is formed.
If the experimentally measured molecular mass of the compound is 153.8 u,
identify the element X.
(Take the average atomic mass of Cl correctly into account.)
Page 2
© RANSHO
