A) 0.6 g
B) 1.0 g
C) 1.5 g
D) 2.0 g
Correct Answer: B
Solution :
[b] (I) Phenopthalein indicate partial neutralisation of \[N{{a}_{2}}C{{O}_{3}}\to NaHC{{O}_{3}}\] |
Meq. of \[N{{a}_{2}}C{{O}_{3}}\]+ Meq. of NaOH = Meq. of HCl |
\[\frac{W}{E}\times 1000+\frac{W}{E}\times 1000=NV\] |
(Suppose \[N{{a}_{2}}C{{O}_{3}}=a\,gm\], NaOH = b gm) |
\[\frac{a}{106}\times 1000+\frac{b}{40}\times 1000=300\times 0.1\].....(1) |
(II) Methyl orange indicate complete neutralisation |
HCl HCl |
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\], \[25\times 0.2=0.1\]\[\times \]\[{{V}_{2}}\,\text{so}\,{{V}_{\text{2}}}=50ml\]excess |
\[\therefore \] \[\frac{a}{53}\times 1000+\frac{b}{40}\times 1000=350\times 0.1\].....(2) |
From (1) and (2) b =1gm. |
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