A) 8.4
B) 15.9
C) 4.0
D) 10.6
Correct Answer: D
Solution :
Molecular weight of \[NaHC{{O}_{3}}=23+1+12+48=48\] 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 \[N{{a}_{2}}C{{O}_{3}}\] is = 106 \[\therefore \] In 19 g of a mixture weight of \[N{{a}_{2}}C{{O}_{3}}=\frac{106\times 19}{190}\] = 10.6gYou need to login to perform this action.
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