4 Bromide
##Topic: Principles Involved in Titrimetric
Exerices, Calorimetry, Preparation of Colloidal Sols and Kinetic Study of Reaction Between \(\mathrm{I}^{-}\)and \(\mathrm{H}_{2} \mathrm{O}_{2}\)
Explanation:
(b) Brown ring test is done for the confirmation of \(\mathrm{NO}_{3}^{-}\)ions.
NOTES \(\begin{aligned} & \mathrm{NO}_{3}^{-}+3 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \rightarrow \mathrm{NO}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)^{6}\right]^{2+}+\mathrm{NO} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2+}+\mathrm{H}_{2} \mathrm{O}}\end{aligned}\) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)^{6}\right]^{2+}+\mathrm{NO} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2+}+\mathrm{H}_{2} \mathrm{O}\)
[Brown ring]
PRACTICAL CHEMISTRY
286938
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
1 750 mg
2 250 mg
3 Zero mg
4 200 mg
Explanation:
(b) \(\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\)
Thus, 1 mole of NaOH reacts with 1 mole of HCl .
Now, Number of moles present are
\(\mathrm{NaOH}=\frac{1 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}=0.025 \mathrm{~mol}\)
\(\mathrm{HCl}=0.025 \mathrm{~L} \times 0.75 \mathrm{M}=0.01875 \mathrm{~mol}\).
\(\Rightarrow\) Number of moles of NaOH left \(=\) Number
of moles of NaOH present - Number of moles of HCl present (limiting reagent)
\(=0.025-0.01875=0.00625 \mathrm{~mol}\)
\(\Rightarrow\) Mass of NaOH left \(=0.00625 \times 40\)
\(=0.25 \mathrm{~g}=250 \mathrm{mg}\)
4 Bromide
##Topic: Principles Involved in Titrimetric
Exerices, Calorimetry, Preparation of Colloidal Sols and Kinetic Study of Reaction Between \(\mathrm{I}^{-}\)and \(\mathrm{H}_{2} \mathrm{O}_{2}\)
Explanation:
(b) Brown ring test is done for the confirmation of \(\mathrm{NO}_{3}^{-}\)ions.
NOTES \(\begin{aligned} & \mathrm{NO}_{3}^{-}+3 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \rightarrow \mathrm{NO}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)^{6}\right]^{2+}+\mathrm{NO} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2+}+\mathrm{H}_{2} \mathrm{O}}\end{aligned}\) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)^{6}\right]^{2+}+\mathrm{NO} \rightarrow\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2+}+\mathrm{H}_{2} \mathrm{O}\)
[Brown ring]
PRACTICAL CHEMISTRY
286938
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
1 750 mg
2 250 mg
3 Zero mg
4 200 mg
Explanation:
(b) \(\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\)
Thus, 1 mole of NaOH reacts with 1 mole of HCl .
Now, Number of moles present are
\(\mathrm{NaOH}=\frac{1 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}=0.025 \mathrm{~mol}\)
\(\mathrm{HCl}=0.025 \mathrm{~L} \times 0.75 \mathrm{M}=0.01875 \mathrm{~mol}\).
\(\Rightarrow\) Number of moles of NaOH left \(=\) Number
of moles of NaOH present - Number of moles of HCl present (limiting reagent)
\(=0.025-0.01875=0.00625 \mathrm{~mol}\)
\(\Rightarrow\) Mass of NaOH left \(=0.00625 \times 40\)
\(=0.25 \mathrm{~g}=250 \mathrm{mg}\)
