A) 8.4
B) 15.9
C) 4.0
D) 1.6
Correct Answer: D
Solution :
Molecular weight of \[NaHC{{O}_{3}}=23+1+12+48=84\] Molecular weight of \[N{{a}_{2}}C{{O}_{3}}=46+12+48=106\] Hence, total weight \[=84+106=190\] \[\because \]In 190 g of a mixture, weight of \[NaC{{O}_{3}}\] is =106 \[\therefore \] In 19 g of a mixture weight of \[N{{a}_{2}}C{{O}_{3}}=\frac{106\times 19}{190}\] \[=10.6g\]You need to login to perform this action.
You will be redirected in
3 sec