A) Sodium carbonate
B) Methyl alcohol
C) Both [a] and [b]
D) Data insufficient
Correct Answer: B
Solution :
Gram formula weight or weight of \[1\] mole of \[N{{a}_{2}}C{{O}_{3}}=106\,\,g\] Gram formula weight of methyl alcohol or weight of one mole of\[C{{H}_{3}}OH=32\,\,g\] Number of moles of\[N{{a}_{2}}C{{O}_{3}}\]produced per month\[=\frac{424\times {{10}^{6}}}{106}=4\times {{10}^{6}}\]moles. The number of moles of \[C{{H}_{3}}OH\]produced per month\[=\frac{320\times {{10}^{6}}}{32}=10\times {{10}^{6}}\]moles. Hence, methyl alcohol is produced more than sodium carbonate in terms of number of moles.You need to login to perform this action.
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