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, 16:22: PM
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](https://images.topperlearning.com/topper/tinymce/cache/bd2b9a033dc04daf0b3bf439928446ca.png)
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, 16:01: PM
CBSE 12-science - Chemistry
Asked by ayazanwarneet | 03 Jan, 2023, 22:58: PM
CBSE 12-science - Chemistry
Asked by gopirammeghwal78 | 05 Aug, 2021, 08:18: AM
CBSE 12-science - Chemistry
Asked by msambyal307 | 30 May, 2021, 14:03: PM
CBSE 12-science - Chemistry
Asked by nishchaymakhija115 | 11 Sep, 2019, 14:31: PM
CBSE 12-science - Chemistry
Asked by pandeyn1604 | 29 May, 2019, 09:16: AM
CBSE 12-science - Chemistry
Asked by rakeshraghav33 | 18 Sep, 2018, 16:22: PM
CBSE 12-science - Chemistry
Asked by Avadhut Katkar | 14 Sep, 2018, 12:37: PM
CBSE 12-science - Chemistry
Asked by minipkda | 21 May, 2018, 22:19: PM
CBSE 12-science - Chemistry
Asked by Topperlearning User | 20 Jun, 2016, 16:00: PM