37451
Phenolphthalein is not suitable for the titration of
1 \(NaOH\) \(vs\) \((COOH)_2\)
2 \(KOH\) \(vs\) \(H_2SO_4\)
3 \(K_2CO_3\) \(vs\) \(HCl\)
4 \(None \,of \,these\)
Explanation:
It’s obvious.
PRACTICAL CHEMISTRY
37452
In order to prepare one litre normal solution of \(KMnO_4\), how many gm of \(KMnO_4\) are required, if the solution is to be used in acid medium for oxidation......\(gm\)
1 \(158\)
2 \(31.60\)
3 \(62\)
4 \(790\)
Explanation:
(b) In acidic medium \(2\) molecules of \(KMn{O_4}\) gives \(5\) atoms of oxygen \(2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5(O)\) \(2 \times 158 = \frac{{316 \times 8}}{{80}} = 31.6\). So, equivalent wt. of \(KMn{O_4}\) in acidic medium is = \(31.6\) \(gm\)
PRACTICAL CHEMISTRY
37453
\(20 \,ml\) of a solution of a weak monobasic acid neutralizes \(22.18 \,ml \) of a solution of \(NaOH\) and \(20\, ml\) of \(N/10\) \(HCl \) neutralizes \(21.5 \,ml\) of the same \(NaOH \) solution. The normality for the acid is nearly.......\(N\)
37455
The maximum amount of \(BaS{O_4}\) precipitated on mixing \(BaC{l_2}\) \((0.5\,M)\) with \({H_2}S{O_4}\) \((1\,M)\) will correspond to.....\(M\)
1 \(0.5\)
2 \(1\)
3 \(1.5\)
4 \(2\)
Explanation:
\(BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCl\) \(1 \,mol\) of \(BaCl _2\) reacts with \(1 \,mol H _2 SO _4\) to give \(1 \,mol\) of \(BaSO _4\). Here, \(BaCl _2\) is the limiting reagent. Therefore, moles of \(BaSO _4\) formed $=$ moles of \(BaCl _2\) present \(=0.5\, mol\).
37451
Phenolphthalein is not suitable for the titration of
1 \(NaOH\) \(vs\) \((COOH)_2\)
2 \(KOH\) \(vs\) \(H_2SO_4\)
3 \(K_2CO_3\) \(vs\) \(HCl\)
4 \(None \,of \,these\)
Explanation:
It’s obvious.
PRACTICAL CHEMISTRY
37452
In order to prepare one litre normal solution of \(KMnO_4\), how many gm of \(KMnO_4\) are required, if the solution is to be used in acid medium for oxidation......\(gm\)
1 \(158\)
2 \(31.60\)
3 \(62\)
4 \(790\)
Explanation:
(b) In acidic medium \(2\) molecules of \(KMn{O_4}\) gives \(5\) atoms of oxygen \(2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5(O)\) \(2 \times 158 = \frac{{316 \times 8}}{{80}} = 31.6\). So, equivalent wt. of \(KMn{O_4}\) in acidic medium is = \(31.6\) \(gm\)
PRACTICAL CHEMISTRY
37453
\(20 \,ml\) of a solution of a weak monobasic acid neutralizes \(22.18 \,ml \) of a solution of \(NaOH\) and \(20\, ml\) of \(N/10\) \(HCl \) neutralizes \(21.5 \,ml\) of the same \(NaOH \) solution. The normality for the acid is nearly.......\(N\)
37455
The maximum amount of \(BaS{O_4}\) precipitated on mixing \(BaC{l_2}\) \((0.5\,M)\) with \({H_2}S{O_4}\) \((1\,M)\) will correspond to.....\(M\)
1 \(0.5\)
2 \(1\)
3 \(1.5\)
4 \(2\)
Explanation:
\(BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCl\) \(1 \,mol\) of \(BaCl _2\) reacts with \(1 \,mol H _2 SO _4\) to give \(1 \,mol\) of \(BaSO _4\). Here, \(BaCl _2\) is the limiting reagent. Therefore, moles of \(BaSO _4\) formed $=$ moles of \(BaCl _2\) present \(=0.5\, mol\).
37451
Phenolphthalein is not suitable for the titration of
1 \(NaOH\) \(vs\) \((COOH)_2\)
2 \(KOH\) \(vs\) \(H_2SO_4\)
3 \(K_2CO_3\) \(vs\) \(HCl\)
4 \(None \,of \,these\)
Explanation:
It’s obvious.
PRACTICAL CHEMISTRY
37452
In order to prepare one litre normal solution of \(KMnO_4\), how many gm of \(KMnO_4\) are required, if the solution is to be used in acid medium for oxidation......\(gm\)
1 \(158\)
2 \(31.60\)
3 \(62\)
4 \(790\)
Explanation:
(b) In acidic medium \(2\) molecules of \(KMn{O_4}\) gives \(5\) atoms of oxygen \(2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5(O)\) \(2 \times 158 = \frac{{316 \times 8}}{{80}} = 31.6\). So, equivalent wt. of \(KMn{O_4}\) in acidic medium is = \(31.6\) \(gm\)
PRACTICAL CHEMISTRY
37453
\(20 \,ml\) of a solution of a weak monobasic acid neutralizes \(22.18 \,ml \) of a solution of \(NaOH\) and \(20\, ml\) of \(N/10\) \(HCl \) neutralizes \(21.5 \,ml\) of the same \(NaOH \) solution. The normality for the acid is nearly.......\(N\)
37455
The maximum amount of \(BaS{O_4}\) precipitated on mixing \(BaC{l_2}\) \((0.5\,M)\) with \({H_2}S{O_4}\) \((1\,M)\) will correspond to.....\(M\)
1 \(0.5\)
2 \(1\)
3 \(1.5\)
4 \(2\)
Explanation:
\(BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCl\) \(1 \,mol\) of \(BaCl _2\) reacts with \(1 \,mol H _2 SO _4\) to give \(1 \,mol\) of \(BaSO _4\). Here, \(BaCl _2\) is the limiting reagent. Therefore, moles of \(BaSO _4\) formed $=$ moles of \(BaCl _2\) present \(=0.5\, mol\).
37451
Phenolphthalein is not suitable for the titration of
1 \(NaOH\) \(vs\) \((COOH)_2\)
2 \(KOH\) \(vs\) \(H_2SO_4\)
3 \(K_2CO_3\) \(vs\) \(HCl\)
4 \(None \,of \,these\)
Explanation:
It’s obvious.
PRACTICAL CHEMISTRY
37452
In order to prepare one litre normal solution of \(KMnO_4\), how many gm of \(KMnO_4\) are required, if the solution is to be used in acid medium for oxidation......\(gm\)
1 \(158\)
2 \(31.60\)
3 \(62\)
4 \(790\)
Explanation:
(b) In acidic medium \(2\) molecules of \(KMn{O_4}\) gives \(5\) atoms of oxygen \(2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5(O)\) \(2 \times 158 = \frac{{316 \times 8}}{{80}} = 31.6\). So, equivalent wt. of \(KMn{O_4}\) in acidic medium is = \(31.6\) \(gm\)
PRACTICAL CHEMISTRY
37453
\(20 \,ml\) of a solution of a weak monobasic acid neutralizes \(22.18 \,ml \) of a solution of \(NaOH\) and \(20\, ml\) of \(N/10\) \(HCl \) neutralizes \(21.5 \,ml\) of the same \(NaOH \) solution. The normality for the acid is nearly.......\(N\)
37455
The maximum amount of \(BaS{O_4}\) precipitated on mixing \(BaC{l_2}\) \((0.5\,M)\) with \({H_2}S{O_4}\) \((1\,M)\) will correspond to.....\(M\)
1 \(0.5\)
2 \(1\)
3 \(1.5\)
4 \(2\)
Explanation:
\(BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCl\) \(1 \,mol\) of \(BaCl _2\) reacts with \(1 \,mol H _2 SO _4\) to give \(1 \,mol\) of \(BaSO _4\). Here, \(BaCl _2\) is the limiting reagent. Therefore, moles of \(BaSO _4\) formed $=$ moles of \(BaCl _2\) present \(=0.5\, mol\).
37451
Phenolphthalein is not suitable for the titration of
1 \(NaOH\) \(vs\) \((COOH)_2\)
2 \(KOH\) \(vs\) \(H_2SO_4\)
3 \(K_2CO_3\) \(vs\) \(HCl\)
4 \(None \,of \,these\)
Explanation:
It’s obvious.
PRACTICAL CHEMISTRY
37452
In order to prepare one litre normal solution of \(KMnO_4\), how many gm of \(KMnO_4\) are required, if the solution is to be used in acid medium for oxidation......\(gm\)
1 \(158\)
2 \(31.60\)
3 \(62\)
4 \(790\)
Explanation:
(b) In acidic medium \(2\) molecules of \(KMn{O_4}\) gives \(5\) atoms of oxygen \(2KMn{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + 5(O)\) \(2 \times 158 = \frac{{316 \times 8}}{{80}} = 31.6\). So, equivalent wt. of \(KMn{O_4}\) in acidic medium is = \(31.6\) \(gm\)
PRACTICAL CHEMISTRY
37453
\(20 \,ml\) of a solution of a weak monobasic acid neutralizes \(22.18 \,ml \) of a solution of \(NaOH\) and \(20\, ml\) of \(N/10\) \(HCl \) neutralizes \(21.5 \,ml\) of the same \(NaOH \) solution. The normality for the acid is nearly.......\(N\)
37455
The maximum amount of \(BaS{O_4}\) precipitated on mixing \(BaC{l_2}\) \((0.5\,M)\) with \({H_2}S{O_4}\) \((1\,M)\) will correspond to.....\(M\)
1 \(0.5\)
2 \(1\)
3 \(1.5\)
4 \(2\)
Explanation:
\(BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCl\) \(1 \,mol\) of \(BaCl _2\) reacts with \(1 \,mol H _2 SO _4\) to give \(1 \,mol\) of \(BaSO _4\). Here, \(BaCl _2\) is the limiting reagent. Therefore, moles of \(BaSO _4\) formed $=$ moles of \(BaCl _2\) present \(=0.5\, mol\).
