37464
The equivalent weight of a metal is \(4.0\). The vapour density of its chloride is \(59.25\). Its atomic weight is
1 \(12\)
2 \(8\)
3 \(36\)
4 \(24\)
Explanation:
Given, equivalent weight of metal \(=4.0\) and vapour density of metal chloride \(=59.25\). \(\therefore\) molecular weight of metal chloride, \(=2 \times V . D=2 \times 59.25\) \(=118.5\) \(\therefore\) Valency of metal, \(=\frac{\text { moleculas weight of metal chloride }}{\text { equivalent weight of metal }+35.5}\) \(=\frac{118.5}{4+35.5}=\frac{118.5}{39.5}=3\) Therefore Atomic weight of the metal $=$equivalent weight \(=4 \times 3\) Atomic weight \(=12\)
PRACTICAL CHEMISTRY
37465
Indicator for the titration of \(HCl\) and \(N{a_2}C{O_3}\) would be
1 \({K_4}Fe{(CN)_6}\)
2 \({K_3}Fe{(CN)_6}\)
3 Phenolphthalein
4 Methyl orange
Explanation:
for the titration between hydrochloric acid and sodium carbonate, the indicator used is methylorange. Methyl orange indicator (\(pH\) Range \(3.0-4.4\)) is used for titrations of a strong acid with a strong or weak base.
PRACTICAL CHEMISTRY
37466
\(20 \,ml\) of a \(N\) solution of \(KMnO_4\) just reacts with \(20 \,ml\) of a solution of oxalic acid. The weight of oxalic acid crystals in \(1\,N\) of the solution is......\(g\)
1 \(31.5\)
2 \(126\)
3 \(63\)
4 \(6.3\)
Explanation:
\(20\, ml\, N - KMnO _4 \cong 20\, ml \,N\)-oxalic acid \(\therefore\) Normality of oxalic acid solution \(=1 \,N\) We know, \(w=\frac{N E V}{1000}=\frac{1 \times 63 \times 1000}{1000}\) \(=63\,g\)
PRACTICAL CHEMISTRY
37467
\(0.53\, gm\) of \(N{a_2}C{O_3}\) has been dissolved in \(100 \,ml\) of a sodium carbonate solution. The normality of the solution will be
1 \(\frac{N}{5}\)
2 \(\frac{N}{2}\)
3 \(\frac{N}{{10}}\)
4 \(N\)
Explanation:
(c) \(N = \frac{{0.53 \times 1000}}{{53 \times 100}} \Rightarrow N = \frac{1}{{10}}\) So normality of the solution will be \(\frac{N}{{10}}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PRACTICAL CHEMISTRY
37464
The equivalent weight of a metal is \(4.0\). The vapour density of its chloride is \(59.25\). Its atomic weight is
1 \(12\)
2 \(8\)
3 \(36\)
4 \(24\)
Explanation:
Given, equivalent weight of metal \(=4.0\) and vapour density of metal chloride \(=59.25\). \(\therefore\) molecular weight of metal chloride, \(=2 \times V . D=2 \times 59.25\) \(=118.5\) \(\therefore\) Valency of metal, \(=\frac{\text { moleculas weight of metal chloride }}{\text { equivalent weight of metal }+35.5}\) \(=\frac{118.5}{4+35.5}=\frac{118.5}{39.5}=3\) Therefore Atomic weight of the metal $=$equivalent weight \(=4 \times 3\) Atomic weight \(=12\)
PRACTICAL CHEMISTRY
37465
Indicator for the titration of \(HCl\) and \(N{a_2}C{O_3}\) would be
1 \({K_4}Fe{(CN)_6}\)
2 \({K_3}Fe{(CN)_6}\)
3 Phenolphthalein
4 Methyl orange
Explanation:
for the titration between hydrochloric acid and sodium carbonate, the indicator used is methylorange. Methyl orange indicator (\(pH\) Range \(3.0-4.4\)) is used for titrations of a strong acid with a strong or weak base.
PRACTICAL CHEMISTRY
37466
\(20 \,ml\) of a \(N\) solution of \(KMnO_4\) just reacts with \(20 \,ml\) of a solution of oxalic acid. The weight of oxalic acid crystals in \(1\,N\) of the solution is......\(g\)
1 \(31.5\)
2 \(126\)
3 \(63\)
4 \(6.3\)
Explanation:
\(20\, ml\, N - KMnO _4 \cong 20\, ml \,N\)-oxalic acid \(\therefore\) Normality of oxalic acid solution \(=1 \,N\) We know, \(w=\frac{N E V}{1000}=\frac{1 \times 63 \times 1000}{1000}\) \(=63\,g\)
PRACTICAL CHEMISTRY
37467
\(0.53\, gm\) of \(N{a_2}C{O_3}\) has been dissolved in \(100 \,ml\) of a sodium carbonate solution. The normality of the solution will be
1 \(\frac{N}{5}\)
2 \(\frac{N}{2}\)
3 \(\frac{N}{{10}}\)
4 \(N\)
Explanation:
(c) \(N = \frac{{0.53 \times 1000}}{{53 \times 100}} \Rightarrow N = \frac{1}{{10}}\) So normality of the solution will be \(\frac{N}{{10}}\).
37464
The equivalent weight of a metal is \(4.0\). The vapour density of its chloride is \(59.25\). Its atomic weight is
1 \(12\)
2 \(8\)
3 \(36\)
4 \(24\)
Explanation:
Given, equivalent weight of metal \(=4.0\) and vapour density of metal chloride \(=59.25\). \(\therefore\) molecular weight of metal chloride, \(=2 \times V . D=2 \times 59.25\) \(=118.5\) \(\therefore\) Valency of metal, \(=\frac{\text { moleculas weight of metal chloride }}{\text { equivalent weight of metal }+35.5}\) \(=\frac{118.5}{4+35.5}=\frac{118.5}{39.5}=3\) Therefore Atomic weight of the metal $=$equivalent weight \(=4 \times 3\) Atomic weight \(=12\)
PRACTICAL CHEMISTRY
37465
Indicator for the titration of \(HCl\) and \(N{a_2}C{O_3}\) would be
1 \({K_4}Fe{(CN)_6}\)
2 \({K_3}Fe{(CN)_6}\)
3 Phenolphthalein
4 Methyl orange
Explanation:
for the titration between hydrochloric acid and sodium carbonate, the indicator used is methylorange. Methyl orange indicator (\(pH\) Range \(3.0-4.4\)) is used for titrations of a strong acid with a strong or weak base.
PRACTICAL CHEMISTRY
37466
\(20 \,ml\) of a \(N\) solution of \(KMnO_4\) just reacts with \(20 \,ml\) of a solution of oxalic acid. The weight of oxalic acid crystals in \(1\,N\) of the solution is......\(g\)
1 \(31.5\)
2 \(126\)
3 \(63\)
4 \(6.3\)
Explanation:
\(20\, ml\, N - KMnO _4 \cong 20\, ml \,N\)-oxalic acid \(\therefore\) Normality of oxalic acid solution \(=1 \,N\) We know, \(w=\frac{N E V}{1000}=\frac{1 \times 63 \times 1000}{1000}\) \(=63\,g\)
PRACTICAL CHEMISTRY
37467
\(0.53\, gm\) of \(N{a_2}C{O_3}\) has been dissolved in \(100 \,ml\) of a sodium carbonate solution. The normality of the solution will be
1 \(\frac{N}{5}\)
2 \(\frac{N}{2}\)
3 \(\frac{N}{{10}}\)
4 \(N\)
Explanation:
(c) \(N = \frac{{0.53 \times 1000}}{{53 \times 100}} \Rightarrow N = \frac{1}{{10}}\) So normality of the solution will be \(\frac{N}{{10}}\).
37464
The equivalent weight of a metal is \(4.0\). The vapour density of its chloride is \(59.25\). Its atomic weight is
1 \(12\)
2 \(8\)
3 \(36\)
4 \(24\)
Explanation:
Given, equivalent weight of metal \(=4.0\) and vapour density of metal chloride \(=59.25\). \(\therefore\) molecular weight of metal chloride, \(=2 \times V . D=2 \times 59.25\) \(=118.5\) \(\therefore\) Valency of metal, \(=\frac{\text { moleculas weight of metal chloride }}{\text { equivalent weight of metal }+35.5}\) \(=\frac{118.5}{4+35.5}=\frac{118.5}{39.5}=3\) Therefore Atomic weight of the metal $=$equivalent weight \(=4 \times 3\) Atomic weight \(=12\)
PRACTICAL CHEMISTRY
37465
Indicator for the titration of \(HCl\) and \(N{a_2}C{O_3}\) would be
1 \({K_4}Fe{(CN)_6}\)
2 \({K_3}Fe{(CN)_6}\)
3 Phenolphthalein
4 Methyl orange
Explanation:
for the titration between hydrochloric acid and sodium carbonate, the indicator used is methylorange. Methyl orange indicator (\(pH\) Range \(3.0-4.4\)) is used for titrations of a strong acid with a strong or weak base.
PRACTICAL CHEMISTRY
37466
\(20 \,ml\) of a \(N\) solution of \(KMnO_4\) just reacts with \(20 \,ml\) of a solution of oxalic acid. The weight of oxalic acid crystals in \(1\,N\) of the solution is......\(g\)
1 \(31.5\)
2 \(126\)
3 \(63\)
4 \(6.3\)
Explanation:
\(20\, ml\, N - KMnO _4 \cong 20\, ml \,N\)-oxalic acid \(\therefore\) Normality of oxalic acid solution \(=1 \,N\) We know, \(w=\frac{N E V}{1000}=\frac{1 \times 63 \times 1000}{1000}\) \(=63\,g\)
PRACTICAL CHEMISTRY
37467
\(0.53\, gm\) of \(N{a_2}C{O_3}\) has been dissolved in \(100 \,ml\) of a sodium carbonate solution. The normality of the solution will be
1 \(\frac{N}{5}\)
2 \(\frac{N}{2}\)
3 \(\frac{N}{{10}}\)
4 \(N\)
Explanation:
(c) \(N = \frac{{0.53 \times 1000}}{{53 \times 100}} \Rightarrow N = \frac{1}{{10}}\) So normality of the solution will be \(\frac{N}{{10}}\).