19955
\(2.5\,F\) of electricity are passed through a \(CuS{O_4}\) solution. The number of gm equivalent of \(Cu\) deposited on anode is $....$
1 \(0\)
2 \(1.25\)
3 \(2.5\)
4 \(5.0\)
Explanation:
(c) \(1\,F\) obtained from \(1\,g\) equivalent \(\therefore \) \(2.5\,F\) obtained from \(2.5\,g\) equivalent.
ELECTROCHEMISTRY
19956
The equivalent weight of a certain trivalent element is \(20. \) Molecular weight of its oxide is
1 \(152\)
2 \(56\)
3 \(168\)
4 \(68\)
Explanation:
As the valency is \(3,\) the oxide will be \(M 2 O 3\). If 3 atoms of oxygen are used then they are 6 equivalents. So, 6 equivalents of the element will also be used, molecular weight \(=6^{*} 20+6^{*} 8=120+48=168 .\)
ELECTROCHEMISTRY
19957
Silver is removed electrically from \(200\,ml\) of a \(0.1\,N\) solution of \(AgN{O_3}\) by a current of \( 0.1\) ampere. How long will it take to remove half of the silver from the solution .............. \(\mathrm{sec}\)
1 \(16\)
2 \(9650\)
3 \(100\)
4 \(10\)
Explanation:
Mmoles \(A g N O_{3}=0.1 \times 200=20\) \(A g^{\oplus}+e^{-} \rightarrow A g\) For \(50\, \%\) electrolysis, \(10 \,mF\) is required \(\Rightarrow 10 \,m F=\frac{1\, t}{96500} \Rightarrow t=9650\, s\)
ELECTROCHEMISTRY
19959
A current of \(0.25\,A\) is passed through \(CuS{O_4}\) solution placed in voltameter for \(45 \) minutes. The amount of \(Cu\) deposited on cathode is ............. \(\mathrm{g}\) (At weight of \(Cu = 63.6\))
1 \(0.22\)
2 \(0.20\)
3 \(0.25\)
4 \(0.30\)
Explanation:
Given :- Time \(=45 min =45 \times 60 sec =2700\,sec\) Current\(=0.25\,A\) \(C u^{2+}+2 e^{-} \rightarrow C u\) At.weight of \(Cu =63.6\) Electricity passed \(=45 \times 60 \times 0.25=675\, C\) electricity deposit copper \(=x \times 2 \times 96500\, C\) of electricity will \(M=Z \times I \times T\) \(x=0.20 \,g\)
ELECTROCHEMISTRY
19960
If \(0.5\,amp\) current is passed through acidified silver nitrate solution for \(10 \) minutes. The mass of silver deposited on cathode, is ........... \(\mathrm{g}\) (eq. wt. of silver nitrate \(= 108\))
19955
\(2.5\,F\) of electricity are passed through a \(CuS{O_4}\) solution. The number of gm equivalent of \(Cu\) deposited on anode is $....$
1 \(0\)
2 \(1.25\)
3 \(2.5\)
4 \(5.0\)
Explanation:
(c) \(1\,F\) obtained from \(1\,g\) equivalent \(\therefore \) \(2.5\,F\) obtained from \(2.5\,g\) equivalent.
ELECTROCHEMISTRY
19956
The equivalent weight of a certain trivalent element is \(20. \) Molecular weight of its oxide is
1 \(152\)
2 \(56\)
3 \(168\)
4 \(68\)
Explanation:
As the valency is \(3,\) the oxide will be \(M 2 O 3\). If 3 atoms of oxygen are used then they are 6 equivalents. So, 6 equivalents of the element will also be used, molecular weight \(=6^{*} 20+6^{*} 8=120+48=168 .\)
ELECTROCHEMISTRY
19957
Silver is removed electrically from \(200\,ml\) of a \(0.1\,N\) solution of \(AgN{O_3}\) by a current of \( 0.1\) ampere. How long will it take to remove half of the silver from the solution .............. \(\mathrm{sec}\)
1 \(16\)
2 \(9650\)
3 \(100\)
4 \(10\)
Explanation:
Mmoles \(A g N O_{3}=0.1 \times 200=20\) \(A g^{\oplus}+e^{-} \rightarrow A g\) For \(50\, \%\) electrolysis, \(10 \,mF\) is required \(\Rightarrow 10 \,m F=\frac{1\, t}{96500} \Rightarrow t=9650\, s\)
ELECTROCHEMISTRY
19959
A current of \(0.25\,A\) is passed through \(CuS{O_4}\) solution placed in voltameter for \(45 \) minutes. The amount of \(Cu\) deposited on cathode is ............. \(\mathrm{g}\) (At weight of \(Cu = 63.6\))
1 \(0.22\)
2 \(0.20\)
3 \(0.25\)
4 \(0.30\)
Explanation:
Given :- Time \(=45 min =45 \times 60 sec =2700\,sec\) Current\(=0.25\,A\) \(C u^{2+}+2 e^{-} \rightarrow C u\) At.weight of \(Cu =63.6\) Electricity passed \(=45 \times 60 \times 0.25=675\, C\) electricity deposit copper \(=x \times 2 \times 96500\, C\) of electricity will \(M=Z \times I \times T\) \(x=0.20 \,g\)
ELECTROCHEMISTRY
19960
If \(0.5\,amp\) current is passed through acidified silver nitrate solution for \(10 \) minutes. The mass of silver deposited on cathode, is ........... \(\mathrm{g}\) (eq. wt. of silver nitrate \(= 108\))
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
ELECTROCHEMISTRY
19955
\(2.5\,F\) of electricity are passed through a \(CuS{O_4}\) solution. The number of gm equivalent of \(Cu\) deposited on anode is $....$
1 \(0\)
2 \(1.25\)
3 \(2.5\)
4 \(5.0\)
Explanation:
(c) \(1\,F\) obtained from \(1\,g\) equivalent \(\therefore \) \(2.5\,F\) obtained from \(2.5\,g\) equivalent.
ELECTROCHEMISTRY
19956
The equivalent weight of a certain trivalent element is \(20. \) Molecular weight of its oxide is
1 \(152\)
2 \(56\)
3 \(168\)
4 \(68\)
Explanation:
As the valency is \(3,\) the oxide will be \(M 2 O 3\). If 3 atoms of oxygen are used then they are 6 equivalents. So, 6 equivalents of the element will also be used, molecular weight \(=6^{*} 20+6^{*} 8=120+48=168 .\)
ELECTROCHEMISTRY
19957
Silver is removed electrically from \(200\,ml\) of a \(0.1\,N\) solution of \(AgN{O_3}\) by a current of \( 0.1\) ampere. How long will it take to remove half of the silver from the solution .............. \(\mathrm{sec}\)
1 \(16\)
2 \(9650\)
3 \(100\)
4 \(10\)
Explanation:
Mmoles \(A g N O_{3}=0.1 \times 200=20\) \(A g^{\oplus}+e^{-} \rightarrow A g\) For \(50\, \%\) electrolysis, \(10 \,mF\) is required \(\Rightarrow 10 \,m F=\frac{1\, t}{96500} \Rightarrow t=9650\, s\)
ELECTROCHEMISTRY
19959
A current of \(0.25\,A\) is passed through \(CuS{O_4}\) solution placed in voltameter for \(45 \) minutes. The amount of \(Cu\) deposited on cathode is ............. \(\mathrm{g}\) (At weight of \(Cu = 63.6\))
1 \(0.22\)
2 \(0.20\)
3 \(0.25\)
4 \(0.30\)
Explanation:
Given :- Time \(=45 min =45 \times 60 sec =2700\,sec\) Current\(=0.25\,A\) \(C u^{2+}+2 e^{-} \rightarrow C u\) At.weight of \(Cu =63.6\) Electricity passed \(=45 \times 60 \times 0.25=675\, C\) electricity deposit copper \(=x \times 2 \times 96500\, C\) of electricity will \(M=Z \times I \times T\) \(x=0.20 \,g\)
ELECTROCHEMISTRY
19960
If \(0.5\,amp\) current is passed through acidified silver nitrate solution for \(10 \) minutes. The mass of silver deposited on cathode, is ........... \(\mathrm{g}\) (eq. wt. of silver nitrate \(= 108\))
19955
\(2.5\,F\) of electricity are passed through a \(CuS{O_4}\) solution. The number of gm equivalent of \(Cu\) deposited on anode is $....$
1 \(0\)
2 \(1.25\)
3 \(2.5\)
4 \(5.0\)
Explanation:
(c) \(1\,F\) obtained from \(1\,g\) equivalent \(\therefore \) \(2.5\,F\) obtained from \(2.5\,g\) equivalent.
ELECTROCHEMISTRY
19956
The equivalent weight of a certain trivalent element is \(20. \) Molecular weight of its oxide is
1 \(152\)
2 \(56\)
3 \(168\)
4 \(68\)
Explanation:
As the valency is \(3,\) the oxide will be \(M 2 O 3\). If 3 atoms of oxygen are used then they are 6 equivalents. So, 6 equivalents of the element will also be used, molecular weight \(=6^{*} 20+6^{*} 8=120+48=168 .\)
ELECTROCHEMISTRY
19957
Silver is removed electrically from \(200\,ml\) of a \(0.1\,N\) solution of \(AgN{O_3}\) by a current of \( 0.1\) ampere. How long will it take to remove half of the silver from the solution .............. \(\mathrm{sec}\)
1 \(16\)
2 \(9650\)
3 \(100\)
4 \(10\)
Explanation:
Mmoles \(A g N O_{3}=0.1 \times 200=20\) \(A g^{\oplus}+e^{-} \rightarrow A g\) For \(50\, \%\) electrolysis, \(10 \,mF\) is required \(\Rightarrow 10 \,m F=\frac{1\, t}{96500} \Rightarrow t=9650\, s\)
ELECTROCHEMISTRY
19959
A current of \(0.25\,A\) is passed through \(CuS{O_4}\) solution placed in voltameter for \(45 \) minutes. The amount of \(Cu\) deposited on cathode is ............. \(\mathrm{g}\) (At weight of \(Cu = 63.6\))
1 \(0.22\)
2 \(0.20\)
3 \(0.25\)
4 \(0.30\)
Explanation:
Given :- Time \(=45 min =45 \times 60 sec =2700\,sec\) Current\(=0.25\,A\) \(C u^{2+}+2 e^{-} \rightarrow C u\) At.weight of \(Cu =63.6\) Electricity passed \(=45 \times 60 \times 0.25=675\, C\) electricity deposit copper \(=x \times 2 \times 96500\, C\) of electricity will \(M=Z \times I \times T\) \(x=0.20 \,g\)
ELECTROCHEMISTRY
19960
If \(0.5\,amp\) current is passed through acidified silver nitrate solution for \(10 \) minutes. The mass of silver deposited on cathode, is ........... \(\mathrm{g}\) (eq. wt. of silver nitrate \(= 108\))
19955
\(2.5\,F\) of electricity are passed through a \(CuS{O_4}\) solution. The number of gm equivalent of \(Cu\) deposited on anode is $....$
1 \(0\)
2 \(1.25\)
3 \(2.5\)
4 \(5.0\)
Explanation:
(c) \(1\,F\) obtained from \(1\,g\) equivalent \(\therefore \) \(2.5\,F\) obtained from \(2.5\,g\) equivalent.
ELECTROCHEMISTRY
19956
The equivalent weight of a certain trivalent element is \(20. \) Molecular weight of its oxide is
1 \(152\)
2 \(56\)
3 \(168\)
4 \(68\)
Explanation:
As the valency is \(3,\) the oxide will be \(M 2 O 3\). If 3 atoms of oxygen are used then they are 6 equivalents. So, 6 equivalents of the element will also be used, molecular weight \(=6^{*} 20+6^{*} 8=120+48=168 .\)
ELECTROCHEMISTRY
19957
Silver is removed electrically from \(200\,ml\) of a \(0.1\,N\) solution of \(AgN{O_3}\) by a current of \( 0.1\) ampere. How long will it take to remove half of the silver from the solution .............. \(\mathrm{sec}\)
1 \(16\)
2 \(9650\)
3 \(100\)
4 \(10\)
Explanation:
Mmoles \(A g N O_{3}=0.1 \times 200=20\) \(A g^{\oplus}+e^{-} \rightarrow A g\) For \(50\, \%\) electrolysis, \(10 \,mF\) is required \(\Rightarrow 10 \,m F=\frac{1\, t}{96500} \Rightarrow t=9650\, s\)
ELECTROCHEMISTRY
19959
A current of \(0.25\,A\) is passed through \(CuS{O_4}\) solution placed in voltameter for \(45 \) minutes. The amount of \(Cu\) deposited on cathode is ............. \(\mathrm{g}\) (At weight of \(Cu = 63.6\))
1 \(0.22\)
2 \(0.20\)
3 \(0.25\)
4 \(0.30\)
Explanation:
Given :- Time \(=45 min =45 \times 60 sec =2700\,sec\) Current\(=0.25\,A\) \(C u^{2+}+2 e^{-} \rightarrow C u\) At.weight of \(Cu =63.6\) Electricity passed \(=45 \times 60 \times 0.25=675\, C\) electricity deposit copper \(=x \times 2 \times 96500\, C\) of electricity will \(M=Z \times I \times T\) \(x=0.20 \,g\)
ELECTROCHEMISTRY
19960
If \(0.5\,amp\) current is passed through acidified silver nitrate solution for \(10 \) minutes. The mass of silver deposited on cathode, is ........... \(\mathrm{g}\) (eq. wt. of silver nitrate \(= 108\))