Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 12-science Answered

how many ml of 0.1M HCL are required to react completely with 1g mixture of and NaHCO3 containing equimolar amounts of both?
Asked by rakeshraghav33 | 18 Sep, 2018, 04:22: PM
answered-by-expert Expert Answer
Question:
 
How many ml of 0.1M HCL are required to react completely with a 1g mixture of Na2CO3  and NaHCO3 containing equimolar amounts of both?
 
Solution:
 
Molarity of HCl = 0.1 M Na2CO3 and NaHCO3 = 1 g
 
 
 
The weight of mixture of 
 
 
Let comma space weight space of space Na subscript 2 CO subscript 3 space present space in space mixture space equals space straight x space straight g

therefore space space weight space of space NaHCO subscript 3 space present space in space mixture space equals space open parentheses 1 minus straight x close parentheses space straight g

Molecular space weight space of space Na subscript 2 CO subscript 3 space equals 106 space straight g divided by mol

Molecular space weight space of space NaHCO subscript 3 space equals space 84 space straight g divided by mol

No. space of space moles space of space Na subscript 2 CO subscript 3 space equals straight x over 106

No. space of space moles space of space NaHCO subscript 3 space equals fraction numerator 1 minus straight x over denominator 84 end fraction

As space mixture space of space Na subscript 2 CO subscript 3 space and space NaHCO subscript 3 space containing space equimolar space amounts space of space both comma

straight x over 106 equals fraction numerator 1 minus straight x over denominator 84 end fraction

84 straight x space equals 106 minus 106 straight x

84 straight x space plus space 106 straight x space equals space 106

190 straight x space equals space 106

straight x space equals 106 over 190

space space space equals 0.5578 space straight g

Weight space of space Na subscript 2 CO subscript 3 space present space in space mixture space equals 0.5578 space straight g

therefore space No. space of space moles space of space Na subscript 2 CO subscript 3 space equals fraction numerator 0.5578 over denominator 106 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles

space Weight space of space NaHCO subscript 3 space present space in space mixture space equals space 1 minus 0.5578

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.4422 space straight g


No. space of space moles space of space NaHCO subscript 3 space equals fraction numerator 0.4422 over denominator 84 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles

Now comma

Na subscript 2 CO subscript 3 space plus space 2 HCl space rightwards arrow space 2 NaCl space space plus space straight H subscript 2 straight O space plus space CO subscript 2

space NaHCO subscript 3 space plus space HCl space rightwards arrow space NaCl space plus space straight H subscript 2 straight O space plus space CO subscript 2

So comma space 1 space mole space of space Na subscript 2 CO subscript 3 space requires space 2 space moles space of space HCl

therefore 0.00526 space moles space of space Na subscript 2 CO subscript 3 space will space require space fraction numerator 0.00526 cross times 2 over denominator 1 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.0105 space moles

1 space mole space of space space NaHCO subscript 3 space requires space 1 space moles space of space HCl

therefore space space 0.00526 space moles space of space NaHCO subscript 3 space will space require space space fraction numerator 1 cross times 0.00526 over denominator 1 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00526 space moles

Tota semicolon space HCl space required space equals space 0.0105 plus space 0.00526

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0.01578 space moles space space space space space space space space space space space space space space space space space space space space space space space space space space space space


0.1 space straight M space HCl space means

space 0.1 space mole space of space HCl space in space 1000 space ml

therefore space 0.01578 space moles space of space HCl space will space present space in space space fraction numerator 0.01578 cross times 1000 over denominator 0.1 end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 157.88 space ml
The volume HCl required is 157.88 ml.
Answered by Varsha | 19 Sep, 2018, 11:43: AM
CBSE 12-science - Chemistry
Asked by sagarmishra | 27 Feb, 2024, 04:01: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by ayazanwarneet | 03 Jan, 2023, 10:58: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by gopirammeghwal78 | 05 Aug, 2021, 08:18: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by msambyal307 | 30 May, 2021, 02:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by nishchaymakhija115 | 11 Sep, 2019, 02:31: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by rakeshraghav33 | 18 Sep, 2018, 04:22: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by Avadhut Katkar | 14 Sep, 2018, 12:37: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by minipkda | 21 May, 2018, 10:19: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Chemistry
Asked by Topperlearning User | 20 Jun, 2016, 04:00: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×