A) 0.6 g
B) 1.0 g
C) 1.5 g
D) 2.0 g
Correct Answer: B
Solution :
(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.You need to login to perform this action.
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