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Chemistry Part-I

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How many mL of 0.1 M HCl are required to react completely with

Chemistry Part-I

Class 12

NCERT

Chapter 1: The Solid State

31 questions

Chapter 2: Solutions

54 questions

A solution of glucose in water is labelled as

10

% w/v, what would be the molality and mole fraction of each component in the solution? If the density of the solution is

1.2 g m L^{- 1}

, then what shall be the molarity of the solution?

How many mL of

0.1 M H Cl

are required to react completely with

1 g

mixture of

N a_{2} C O_{3}

and

N a H C O_{3}

containing equimolar amounts of both?

A solution is obtained by mixing

300 g

25

% solution and

400 g

40

% solution by mass. Calculate the mass percentage of the resulting solution?

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Chapter 3: Electrochemistry

28 questions

Chapter 4: Chemical Kinetics

40 questions

Chapter 5: Surface Chemistry

27 questions

Chapter 6: General Principles and Processes of Isolation of Elements

30 questions

Chapter 7: The p-Block Elements

62 questions

Chapter 8: The d-and f-Block Elements

48 questions

Chapter 9: Coordination Compounds

38 questions

Question

Medium

Solving time: 2 mins

How many mL of $0.1 M H Cl$ are required to react completely with $1 g$ mixture of $N a_{2} C O_{3}$ and $N a H C O_{3}$ containing equimolar amounts of both?

Views: 6,546 students

Updated on: Feb 13, 2024

Book:

Chemistry Part-I (NCERT)

Found 5 tutors discussing this question

Charlotte Discussed

How many mL of

0.1 M H Cl

are required to react completely with

1 g

mixture of

N a_{2} C O_{3}

and

N a H C O_{3}

containing equimolar amounts of both?

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

Let the mixture contains

x

g of sodium carbonate and

1 - x

g of sodium bicarbonate.

The molar masses of sodium carbonate and sodium bicarbonate are

106

g/mol and

84

g/mol respectively

The number of moles of sodium carbonate and sodium bicarbonates are

\frac{x}{106}

and

\frac{1 - x}{84}

respectively.

Since, it is an equilmolar mixture,

\frac{x}{106} = \frac{1 - x}{84}

84 x = 106 - 106 x

190 x = 106

x = 0.5579

Number of moles of sodium carbonate

= \frac{0.5579}{106} = 0.005263

Number of moles of sodium hydrogen carbonate

= \frac{1 - 0.5579}{84} = 0.005263

One mole of sodium carbonate will react with

2

moles of

H Cl

and

1

mole of sodium bicarbonate will react with

1

mole of

H Cl

Total number of moles of

H Cl

that will completely neutralize the mixture

= 2 \times 0.005263 + 0.005263 = 0.01578

moles

Volume of

0.1 M H Cl

required

= \frac{0.01578}{0.1} = 0.158 L = 158

mL.

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Practice more questions from Chemistry Part-I (NCERT)

A solution of glucose in water is labelled as

10

% w/v, what would be the molality and mole fraction of each component in the solution? If the density of the solution is

1.2 g m L^{- 1}

, then what shall be the molarity of the solution?

How many mL of

0.1 M H Cl

are required to react completely with

1 g

mixture of

N a_{2} C O_{3}

and

N a H C O_{3}

containing equimolar amounts of both?

A solution is obtained by mixing

300 g

25

% solution and

400 g

40

% solution by mass. Calculate the mass percentage of the resulting solution?

View all

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Question Text	How many mL of $0.1 M H Cl$ are required to react completely with $1 g$ mixture of $N a_{2} C O_{3}$ and $N a H C O_{3}$ containing equimolar amounts of both?
Updated On	Feb 13, 2024
Topic	Solutions
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 19
Upvotes	1912
Avg. Video Duration	10 min

How many mL of $0.1 M H Cl$ are required to react completely with $1 g$ mixture of $N a_{2} C O_{3}$ and $N a H C O_{3}$ containing equimolar amounts of both?

Text solutionVerified

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