Correct Answer - B
No. of moles of `na_(2)CO_(3)` in the residue `=(0.14 xx 1)/(2) = 0.07`
weight loss during heating is due to `CO_(2)` & `H_(2)O`
formed as a result of decomposition of `NaHCO_(3)`
`:.` No. of moles of `Na_(2)CO_(3)` formed due to `NaHCO_(3) = 0.01`
No. of moles of `NaHCO_(3)` in the original mixture `=0.02`
Mass of `NaCI`
`=[10-(0.06 xx 106) +0.02 xx 84] = 1.196`
% by mass of `NaCI = (1.96)/(10) xx 100 = 19.6%`