JEE Exam > JEE Questions > The minimum volume of water required to disso...
Start Learning for Free
The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:
- a)018 L
- b)0.36 L
- c)17.98 L
- d)1.798 L
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The minimum volume of water required to dissolve 0.1g lead (II) chlori...
Ksp of PbCl2 is 3.2 × 10-8
PbCl2 is 3.2 × 10−8
Pbcl2(s) ⇌ Pb2+(aq) + 2Cl−(aq)
t = 0 1 0 0.
At equilibrium 1 – S S 2S.
Ksp = [S] [2S]2
3.2 × 10−8 = 4s3
S3 = 0.8 × 10−8
S3 = 8 × 10−9
S = 2 × 10−3
Solubility = W/V
∴ Solubility of PbCl2 in gL−1 = 2 × 10−3 × 278
= 556 × 10−3gL−1
0.556 gL−1
x = 0.1/0.556
= 0.18L
PbCl2 is 3.2 × 10−8
Pbcl2(s) ⇌ Pb2+(aq) + 2Cl−(aq)
t = 0 1 0 0.
At equilibrium 1 – S S 2S.
Ksp = [S] [2S]2
3.2 × 10−8 = 4s3
S3 = 0.8 × 10−8
S3 = 8 × 10−9
S = 2 × 10−3
Solubility = W/V
∴ Solubility of PbCl2 in gL−1 = 2 × 10−3 × 278
= 556 × 10−3gL−1
0.556 gL−1
0.556/0.1 = 1/x
x = 0.1/0.556
= 0.18L
Most Upvoted Answer
The minimum volume of water required to dissolve 0.1g lead (II) chlori...
To determine the minimum volume of water required to dissolve 0.1g of lead (II) chloride (PbCl2) and obtain a saturated solution, we can use the solubility product constant (Ksp) of PbCl2.
Given data:
Mass of PbCl2 = 0.1g
Ksp of PbCl2 = 3.2 x 10^-8
Atomic mass of Pb = 207 u
To solve this problem, we need to follow these steps:
1. Calculate the number of moles of PbCl2:
Number of moles = Mass / Molar mass
Number of moles of PbCl2 = 0.1g / (207g/mol + 2 x 35.5g/mol) = 0.1g / 278g/mol = 0.000359 moles
2. Write the balanced chemical equation for the dissociation of PbCl2:
PbCl2 (s) ⇌ Pb2+ (aq) + 2Cl- (aq)
3. Use the stoichiometry of the balanced equation to determine the concentration of ions in the saturated solution:
Pb2+ concentration = 0.000359 moles / V (L)
Cl- concentration = 2 x 0.000359 moles / V (L)
4. Apply the solubility product constant expression for PbCl2:
Ksp = [Pb2+] x [Cl-]^2
Substituting the concentrations from step 3:
3.2 x 10^-8 = (0.000359 / V) x (2 x 0.000359 / V)^2
Simplifying the equation:
3.2 x 10^-8 = 4 x (0.000359 / V)^3
(0.000359 / V)^3 = (3.2 x 10^-8) / 4
(0.000359 / V)^3 = 8 x 10^-9
Taking the cube root of both sides:
0.000359 / V = 2 x 10^-3
V = 0.000359 / (2 x 10^-3)
V = 0.0001795 L
5. Convert the volume from liters to milliliters:
V = 0.0001795 L x 1000 mL/L = 0.1795 mL
Therefore, the minimum volume of water required to dissolve 0.1g of PbCl2 and obtain a saturated solution is approximately 0.1795 mL, which is equivalent to 0.018 L (option A).
Given data:
Mass of PbCl2 = 0.1g
Ksp of PbCl2 = 3.2 x 10^-8
Atomic mass of Pb = 207 u
To solve this problem, we need to follow these steps:
1. Calculate the number of moles of PbCl2:
Number of moles = Mass / Molar mass
Number of moles of PbCl2 = 0.1g / (207g/mol + 2 x 35.5g/mol) = 0.1g / 278g/mol = 0.000359 moles
2. Write the balanced chemical equation for the dissociation of PbCl2:
PbCl2 (s) ⇌ Pb2+ (aq) + 2Cl- (aq)
3. Use the stoichiometry of the balanced equation to determine the concentration of ions in the saturated solution:
Pb2+ concentration = 0.000359 moles / V (L)
Cl- concentration = 2 x 0.000359 moles / V (L)
4. Apply the solubility product constant expression for PbCl2:
Ksp = [Pb2+] x [Cl-]^2
Substituting the concentrations from step 3:
3.2 x 10^-8 = (0.000359 / V) x (2 x 0.000359 / V)^2
Simplifying the equation:
3.2 x 10^-8 = 4 x (0.000359 / V)^3
(0.000359 / V)^3 = (3.2 x 10^-8) / 4
(0.000359 / V)^3 = 8 x 10^-9
Taking the cube root of both sides:
0.000359 / V = 2 x 10^-3
V = 0.000359 / (2 x 10^-3)
V = 0.0001795 L
5. Convert the volume from liters to milliliters:
V = 0.0001795 L x 1000 mL/L = 0.1795 mL
Therefore, the minimum volume of water required to dissolve 0.1g of PbCl2 and obtain a saturated solution is approximately 0.1795 mL, which is equivalent to 0.018 L (option A).
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer?
Question Description
The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer?.
The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution(Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:a)018 Lb)0.36 Lc)17.98 Ld)1.798 LCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup