One gram equimolecular mixture of `Na_(2)CO_(3)` and `NaHCO_(3)`is reacted with 0.1 NHCl. The milliliters of 0.1 N HCl required to react completely with the above mixture is :

Options

A. 15.78 mL
B. 157.8 mL
C. 198.4 mL
D. 195.5 mL

Answer

B

Explanation

Suppose the molecules of `Na_(2)CO_(3)` and `NaHCO_(3)` in a mixture are 'a'.
milli-equivalent of HCl
`N_(1)V_(1)+N_(2)V_(2)=NV`
`axx2xx1000+axx1xx1000=0.1V`
`3a=10^(-4)V " " ...(i)`
[`:'` N=basicity /acidity `xx`M]
wt. of `Na_(2)CO_(3)+ ` wt. of `NaHCO_(3)=1g`
(`:.` wt. of mixture = 1g)
`rArr a xx106+axx84=1`
`a=5.26xx10^(-3) " " ....(ii)`
From Eqs. (i) and (ii) we have `3xx5.26xx10^(-3)=10^(-4)VrArrV=157.8` mL