JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 20)
Consider the following reactions :
$$NaCl{\rm{ }} + {\rm{ }}{K_2}C{r_2}{O_7} + \mathop {{H_2}S{O_4}}\limits_{(conc.)} $$$$ \to $$ (A) + side products
(A) + NaOH $$ \to $$ (B) + Side products
$$\left( B \right){\rm{ }} + \mathop {{H_2}S{O_4}}\limits_{(dilute)} + {\rm{ }}{H_2}{O_2}$$ $$ \to $$ (C) + Side products
The sum of the total number of atoms in one molecule each of (A) and (B) and (C) is
$$NaCl{\rm{ }} + {\rm{ }}{K_2}C{r_2}{O_7} + \mathop {{H_2}S{O_4}}\limits_{(conc.)} $$$$ \to $$ (A) + side products
(A) + NaOH $$ \to $$ (B) + Side products
$$\left( B \right){\rm{ }} + \mathop {{H_2}S{O_4}}\limits_{(dilute)} + {\rm{ }}{H_2}{O_2}$$ $$ \to $$ (C) + Side products
The sum of the total number of atoms in one molecule each of (A) and (B) and (C) is
Answer
18
Explanation
$$NaCl{\rm{ }} + {\rm{ }}{K_2}C{r_2}{O_7} + \mathop {{H_2}S{O_4}}\limits_{(conc.)} $$$$ \to $$
2CrO2Cl2(A) + 4NaHSO4 + 2KHSO4 + 3H2O
CrO2Cl2(A) + 4NaOH $$ \to $$ Na2CrO4(B) + 2NaCl + 2H2O
Na2CrO4(B) + 2H2SO4 + 2H2O2 $$ \to $$
CrO5(C) + 2NaHSO4 + 3H2O
A = CrO2Cl2
B = Na2CrO4
C = CrO5
Total number of atom in A + B + C = 18
2CrO2Cl2(A) + 4NaHSO4 + 2KHSO4 + 3H2O
CrO2Cl2(A) + 4NaOH $$ \to $$ Na2CrO4(B) + 2NaCl + 2H2O
Na2CrO4(B) + 2H2SO4 + 2H2O2 $$ \to $$
CrO5(C) + 2NaHSO4 + 3H2O
A = CrO2Cl2
B = Na2CrO4
C = CrO5
Total number of atom in A + B + C = 18
Comments (0)
