Chemistry · Metallurgy

JEE Advanced 2019 — Paper 1 — Question 19

Calamine, malachite, magnetite and cryolite, respectively, are

Option A:
$\mathrm{ZnSO}_{4}, \mathrm{Cu}(\mathrm{OH})_{2}, \mathrm{Fe}_{3} \mathrm{O}_{4}, \mathrm{Na}_{3} \mathrm{AlF}_{6}$
Option B:
$\mathrm{ZnCO}_{3}, \mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}, \mathrm{Fe}_{3} \mathrm{O}_{4}, \mathrm{Na}_{3} \mathrm{AlF}_{6}$
Correct
Option C:
$\mathrm{ZnSO}_{4}, \mathrm{CuCO}_{3}, \mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{AlF}_{3}$
Option D:
$\mathrm{ZnCO}_{3}, \mathrm{CuCO}_{3}, \mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{Na}_{3} \mathrm{AlF}_{6}$

Answer: B

Calamine - $\mathrm{ZnCO}_{3}$

Malachite $-\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}$

Magnetite - $\mathrm{Fe}_{3} \mathrm{O}_{4}$

Cryolite - $\mathrm{Na}_{3} \mathrm{AlF}_{6}$

Answer key and solution verified before publishing.

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