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Chemistry Part-I

Equilibrium

What is the minimum volume of water required to dissolve 1 g o

Chemistry Part-I

Class 11

NCERT

Chapter 1: Some Basic Concepts of Chemistry

44 questions

Chapter 2: Structure of Atom

85 questions

Chapter 3: Classification of Elements and Periodicity in Properties

50 questions

Chapter 4: Chemical Bonding and Molecular Structure

44 questions

Chapter 5: States of Matter

28 questions

Chapter 6: Thermodynamics and Thermochemistry

36 questions

Chapter 7: Equilibrium

101 questions

What is the maximum concentration of equimolar solutions of ferrous sulphateand sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide,

K_{s p} = 6.3 \times 10^{- 18}

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate,

K_{s p} = 9.1 \times 10^{- 6}

The concentration of sulphide ion in 0.1M

H Cl

solution saturated with hydrogen sulphide is

1.0 \times 10^{- 19} M

. If 10 mL of this is added to 5 mL of 0.04 M solution of

F e SO_{4}, M n Cl_{2}, Z n Cl_{2}

and

C d Cl_{2}

. In which of these solutions precipitation will take place?

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Solving time: 4 mins

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, $K_{s p} = 9.1 \times 10^{- 6}$ ).

Name: Video solution 1: [Solved] What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp=9.1×10−6). - Chemistry Part-I
Uploaded: 2024-02-24T03:41:46.000Z
Duration: 3 min
Description: Video solution 1: [Solved] What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp=9.1×10−6). - Chemistry Part-I

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Updated on: Feb 24, 2024

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Chemistry Part-I (NCERT)

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What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate,

K_{s p} = 9.1 \times 10^{- 6}

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Text solutionVerified

Let s be the solubility of

C a S O_{4}

C a S O_{4} ⇌ C a^{2 +} S O_{4}^{2 -}

[C a^{2 +}] = [S O_{4}^{2 +}] = s

K_{s p} [C a^{2 +}] [S O_{4}^{2 -}] = s \times s = s^{2} = 9.1 \times 1 0^{- 6}

s = 0.003016 M

The molar mass of

C a S O_{4}

40 + 32 + 64 = 136 g

.
The solubility in g/L is

0.003017 \times 136 = 0.410 g / L

.
Hence, 0.41 g dissolves in 1 L.
1 g will dissolve in

\frac{1}{0.41} = 2.43 L

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Practice more questions from Chemistry Part-I (NCERT)

What is the maximum concentration of equimolar solutions of ferrous sulphateand sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide,

K_{s p} = 6.3 \times 10^{- 18}

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate,

K_{s p} = 9.1 \times 10^{- 6}

The concentration of sulphide ion in 0.1M

H Cl

solution saturated with hydrogen sulphide is

1.0 \times 10^{- 19} M

. If 10 mL of this is added to 5 mL of 0.04 M solution of

F e SO_{4}, M n Cl_{2}, Z n Cl_{2}

and

C d Cl_{2}

. In which of these solutions precipitation will take place?

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Question Text	What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, $K_{s p} = 9.1 \times 10^{- 6}$ ).
Updated On	Feb 24, 2024
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	250
Avg. Video Duration	4 min

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, $K_{s p} = 9.1 \times 10^{- 6}$ ).

Text solutionVerified

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What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp​=9.1×10−6).

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