Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_2CO _3 and NaHCO_3 containing equimolar amounts of both?

2.6    How many mL of 0.1\; M\; HCl are required to react completely with 1 \; g mixture of Na_{2}CO_{3}  and NaHCO_{3}  containing equimolar amounts of both?

Answers (1)

best_answer

Total amount of mixture of Na2CO3 and NaHCO3 = 1 g.

Let the amount of Na2CO3 be x g.

So the amount of NaHCO3  will be equal to (1 - x) g.

  Molar\ mass\ of\ Na_2CO_3 = 106\ ;\ molar\ mass\ of\ NaHCO_3 = 84

Now it is given that it is an equimolar mixture.

So,            Moles of Na2CO3 =  Moles of  NaHCO3.

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

or                            x = 0.558 g 

So                       Moles\ of \ Na_2CO_3 = \frac{0.558}{106} = 0.00526

and                     Moles\ of \ NaHCO_3 = \frac{1 - 0.558}{84} = 0.0053

 

                               

It is clear that for 1 mol of Na2CO3  2 mol of HCl is required, similarly for 1 mol of NaHCO3 1 mol of HCl is required.

So number of moles required of HCl =  2(0.00526) + 0.0053 = 0.01578 mol

It is given that molarity of HCl is 0.1 which means 0.1 mol of HCl in 1l of solution.

Thus required volume :

                                    = \frac{0.01578}{0.1} = 0.1578\ l = 157.8\ mL

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date Sheet 2025: CBSE, UPMSP, MPBSE, Class 10, 12 board exam dates out; BSEB, CGBSE schedules soon
anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

Latest Question