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NEET • CHEMISTRY • PRACTICAL CHEMISTRY

Practical chemistry practice · Mixed difficulty

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91 Easy
91 of 95
Understanding +4 | -1
In qualitative analysis NH4Cl\mathrm{NH}_4 \mathrm{Cl} is added before NH4OH\mathrm{NH}_4 \mathrm{OH}
View answer and explanation
Correct option: A
to decrease [OH−]\left[\mathrm{OH}^{-}\right] conc.
Option A (to decrease [OH−]\left[\mathrm{OH}^{-}\right] conc.) is correct because adding ammonium chloride before ammonium hydroxide decreases the hydroxide ion concentration due to the common ion effect, preventing the precipitation of group III hydroxides.
92 Medium
92 of 95
Remembering +4 | -1
Disodium hydrogen phoshate is used to test:
View answer and explanation
Correct option: A
Mg2+\mathrm{Mg}^{2+}
Option A (Mg2+\mathrm{Mg}^{2+}) is correct because disodium hydrogen phosphate is used as a reagent to test for magnesium ions.
93 Hard
93 of 95
Remembering +4 | -1
Which procedure for dispensing a liquid with
a volumetric pipet is correct?
View answer and explanation
Correct option: C
Draw the liquid above the line on the pipet
using a pipet bulb. With a finger on the top of
the pipet allow the curve of the meniscus to
drop to the line. Place the tip of the pipet
against the side of the receiving container and
allow the liquid to drain
Option C (Draw the liquid above the line...) is correct because it describes the correct procedure for using a volumetric pipet to accurately dispense a liquid.
94 Medium
94 of 95
Understanding +4 | -1
In copper turnings test, pieces of copper is added to a nitrate salt and heated with conc. H2SO4.\mathrm{H}_2 \mathrm{SO}_4. In this reaction copper acts as
View answer and explanation
Correct option: C
Reducing agent
Option C (Reducing agent) is correct because copper loses electrons (oxidation) to reduce nitrate ions to nitrogen dioxide.
95 Medium
95 of 95
Application +4 | -1
At what pH will a 10−3M10^{-3} \mathrm{M} solution of indicator with Kb=10−10K_b=10^{-10} changes colour
View answer and explanation
Correct option: B
4
Option B (4) is correct because the pH at which an indicator changes color is approximately equal to its pKa_a value. Since Kb_b is given, we can calculate pKa_a and thus the pH.
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