19837
If \(96500\) coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of \(0.15\, amperes\) to deposite \(20\,mg\) of copper from a solution of copper sulphate is (Chemical equivalent of copper = \(32\))
19838
How much current should be passed through acidified water for \(100\,s\) to liberate \(0.224\, litre\) of \({H_2}\) ............. \(\mathrm{A}\)
1 \(22.4\)
2 \(19.3\)
3 \(9.65\)
4 \(1\)
Explanation:
(b) \(22.4\) litre \(H_2\) = 1 mole \( H_2\) = \(N\) molecules of \(H_2= 2\,N\) atom of \(H\) So charge required to liberate \(22.4\,litre\) of \(H_2 = 2\,N\) \(e =2\,F\) Hence charge required to liberate \(0.224\,litre\) of \(H_2 \) = \(\frac{{2F}}{{22.4}} \times 0.224\) \( = \frac{{2F}}{{100}} = 2 \times 965\,C\) So current \(i = \frac{Q}{t} = \frac{{2 \times 965}}{{100}} = 19.3\,amp\)
ELECTROCHEMISTRY
19839
If a steady current of \(4\, amp\) maintained for \(40\) minutes, deposits \(4.5\, gm\) of zinc at the cathode and then the electro chemical equivalent will be
19861
The electric conduction of a salt solution in water depends on the
1 Shape of its molecules
2 Size of its molecules
3 Size of solvent molecules
4 Extent of its ionization
Explanation:
Higher the number of ions in solution, higher would be the conductivity of the solution as the number of charge carriers would increase. Hence, higher the degree of ionization greater would be the number of ions in solution.
ELECTROCHEMISTRY
19840
The current flowing in a copper voltameter is \(3.2\, A\). The number of copper ions \((C{u^{2 + }})\) deposited at the cathode per minute is
1 \(0.5 \times {10^{20}}\)
2 \(1.5 \times {10^{20}}\)
3 \(3 \times {10^{20}}\)
4 \(6 \times {10^{20}}\)
Explanation:
(d) Charge supplied per minute = \(3.2 × 60\) = \(192\,C\) Charge \(2e\) liberates one \(Cu^{+2}\) ion No of \(Cu^{+2}\) ion liberate by \(192\,C\) \( = \frac{{192}}{{2e}} = \frac{{192}}{{2 \times 1.6 \times {{10}^{ - 19}}}} = 6 \times {10^{20}}\)
19837
If \(96500\) coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of \(0.15\, amperes\) to deposite \(20\,mg\) of copper from a solution of copper sulphate is (Chemical equivalent of copper = \(32\))
19838
How much current should be passed through acidified water for \(100\,s\) to liberate \(0.224\, litre\) of \({H_2}\) ............. \(\mathrm{A}\)
1 \(22.4\)
2 \(19.3\)
3 \(9.65\)
4 \(1\)
Explanation:
(b) \(22.4\) litre \(H_2\) = 1 mole \( H_2\) = \(N\) molecules of \(H_2= 2\,N\) atom of \(H\) So charge required to liberate \(22.4\,litre\) of \(H_2 = 2\,N\) \(e =2\,F\) Hence charge required to liberate \(0.224\,litre\) of \(H_2 \) = \(\frac{{2F}}{{22.4}} \times 0.224\) \( = \frac{{2F}}{{100}} = 2 \times 965\,C\) So current \(i = \frac{Q}{t} = \frac{{2 \times 965}}{{100}} = 19.3\,amp\)
ELECTROCHEMISTRY
19839
If a steady current of \(4\, amp\) maintained for \(40\) minutes, deposits \(4.5\, gm\) of zinc at the cathode and then the electro chemical equivalent will be
19861
The electric conduction of a salt solution in water depends on the
1 Shape of its molecules
2 Size of its molecules
3 Size of solvent molecules
4 Extent of its ionization
Explanation:
Higher the number of ions in solution, higher would be the conductivity of the solution as the number of charge carriers would increase. Hence, higher the degree of ionization greater would be the number of ions in solution.
ELECTROCHEMISTRY
19840
The current flowing in a copper voltameter is \(3.2\, A\). The number of copper ions \((C{u^{2 + }})\) deposited at the cathode per minute is
1 \(0.5 \times {10^{20}}\)
2 \(1.5 \times {10^{20}}\)
3 \(3 \times {10^{20}}\)
4 \(6 \times {10^{20}}\)
Explanation:
(d) Charge supplied per minute = \(3.2 × 60\) = \(192\,C\) Charge \(2e\) liberates one \(Cu^{+2}\) ion No of \(Cu^{+2}\) ion liberate by \(192\,C\) \( = \frac{{192}}{{2e}} = \frac{{192}}{{2 \times 1.6 \times {{10}^{ - 19}}}} = 6 \times {10^{20}}\)
19837
If \(96500\) coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of \(0.15\, amperes\) to deposite \(20\,mg\) of copper from a solution of copper sulphate is (Chemical equivalent of copper = \(32\))
19838
How much current should be passed through acidified water for \(100\,s\) to liberate \(0.224\, litre\) of \({H_2}\) ............. \(\mathrm{A}\)
1 \(22.4\)
2 \(19.3\)
3 \(9.65\)
4 \(1\)
Explanation:
(b) \(22.4\) litre \(H_2\) = 1 mole \( H_2\) = \(N\) molecules of \(H_2= 2\,N\) atom of \(H\) So charge required to liberate \(22.4\,litre\) of \(H_2 = 2\,N\) \(e =2\,F\) Hence charge required to liberate \(0.224\,litre\) of \(H_2 \) = \(\frac{{2F}}{{22.4}} \times 0.224\) \( = \frac{{2F}}{{100}} = 2 \times 965\,C\) So current \(i = \frac{Q}{t} = \frac{{2 \times 965}}{{100}} = 19.3\,amp\)
ELECTROCHEMISTRY
19839
If a steady current of \(4\, amp\) maintained for \(40\) minutes, deposits \(4.5\, gm\) of zinc at the cathode and then the electro chemical equivalent will be
19861
The electric conduction of a salt solution in water depends on the
1 Shape of its molecules
2 Size of its molecules
3 Size of solvent molecules
4 Extent of its ionization
Explanation:
Higher the number of ions in solution, higher would be the conductivity of the solution as the number of charge carriers would increase. Hence, higher the degree of ionization greater would be the number of ions in solution.
ELECTROCHEMISTRY
19840
The current flowing in a copper voltameter is \(3.2\, A\). The number of copper ions \((C{u^{2 + }})\) deposited at the cathode per minute is
1 \(0.5 \times {10^{20}}\)
2 \(1.5 \times {10^{20}}\)
3 \(3 \times {10^{20}}\)
4 \(6 \times {10^{20}}\)
Explanation:
(d) Charge supplied per minute = \(3.2 × 60\) = \(192\,C\) Charge \(2e\) liberates one \(Cu^{+2}\) ion No of \(Cu^{+2}\) ion liberate by \(192\,C\) \( = \frac{{192}}{{2e}} = \frac{{192}}{{2 \times 1.6 \times {{10}^{ - 19}}}} = 6 \times {10^{20}}\)
19837
If \(96500\) coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of \(0.15\, amperes\) to deposite \(20\,mg\) of copper from a solution of copper sulphate is (Chemical equivalent of copper = \(32\))
19838
How much current should be passed through acidified water for \(100\,s\) to liberate \(0.224\, litre\) of \({H_2}\) ............. \(\mathrm{A}\)
1 \(22.4\)
2 \(19.3\)
3 \(9.65\)
4 \(1\)
Explanation:
(b) \(22.4\) litre \(H_2\) = 1 mole \( H_2\) = \(N\) molecules of \(H_2= 2\,N\) atom of \(H\) So charge required to liberate \(22.4\,litre\) of \(H_2 = 2\,N\) \(e =2\,F\) Hence charge required to liberate \(0.224\,litre\) of \(H_2 \) = \(\frac{{2F}}{{22.4}} \times 0.224\) \( = \frac{{2F}}{{100}} = 2 \times 965\,C\) So current \(i = \frac{Q}{t} = \frac{{2 \times 965}}{{100}} = 19.3\,amp\)
ELECTROCHEMISTRY
19839
If a steady current of \(4\, amp\) maintained for \(40\) minutes, deposits \(4.5\, gm\) of zinc at the cathode and then the electro chemical equivalent will be
19861
The electric conduction of a salt solution in water depends on the
1 Shape of its molecules
2 Size of its molecules
3 Size of solvent molecules
4 Extent of its ionization
Explanation:
Higher the number of ions in solution, higher would be the conductivity of the solution as the number of charge carriers would increase. Hence, higher the degree of ionization greater would be the number of ions in solution.
ELECTROCHEMISTRY
19840
The current flowing in a copper voltameter is \(3.2\, A\). The number of copper ions \((C{u^{2 + }})\) deposited at the cathode per minute is
1 \(0.5 \times {10^{20}}\)
2 \(1.5 \times {10^{20}}\)
3 \(3 \times {10^{20}}\)
4 \(6 \times {10^{20}}\)
Explanation:
(d) Charge supplied per minute = \(3.2 × 60\) = \(192\,C\) Charge \(2e\) liberates one \(Cu^{+2}\) ion No of \(Cu^{+2}\) ion liberate by \(192\,C\) \( = \frac{{192}}{{2e}} = \frac{{192}}{{2 \times 1.6 \times {{10}^{ - 19}}}} = 6 \times {10^{20}}\)
19837
If \(96500\) coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of \(0.15\, amperes\) to deposite \(20\,mg\) of copper from a solution of copper sulphate is (Chemical equivalent of copper = \(32\))
19838
How much current should be passed through acidified water for \(100\,s\) to liberate \(0.224\, litre\) of \({H_2}\) ............. \(\mathrm{A}\)
1 \(22.4\)
2 \(19.3\)
3 \(9.65\)
4 \(1\)
Explanation:
(b) \(22.4\) litre \(H_2\) = 1 mole \( H_2\) = \(N\) molecules of \(H_2= 2\,N\) atom of \(H\) So charge required to liberate \(22.4\,litre\) of \(H_2 = 2\,N\) \(e =2\,F\) Hence charge required to liberate \(0.224\,litre\) of \(H_2 \) = \(\frac{{2F}}{{22.4}} \times 0.224\) \( = \frac{{2F}}{{100}} = 2 \times 965\,C\) So current \(i = \frac{Q}{t} = \frac{{2 \times 965}}{{100}} = 19.3\,amp\)
ELECTROCHEMISTRY
19839
If a steady current of \(4\, amp\) maintained for \(40\) minutes, deposits \(4.5\, gm\) of zinc at the cathode and then the electro chemical equivalent will be
19861
The electric conduction of a salt solution in water depends on the
1 Shape of its molecules
2 Size of its molecules
3 Size of solvent molecules
4 Extent of its ionization
Explanation:
Higher the number of ions in solution, higher would be the conductivity of the solution as the number of charge carriers would increase. Hence, higher the degree of ionization greater would be the number of ions in solution.
ELECTROCHEMISTRY
19840
The current flowing in a copper voltameter is \(3.2\, A\). The number of copper ions \((C{u^{2 + }})\) deposited at the cathode per minute is
1 \(0.5 \times {10^{20}}\)
2 \(1.5 \times {10^{20}}\)
3 \(3 \times {10^{20}}\)
4 \(6 \times {10^{20}}\)
Explanation:
(d) Charge supplied per minute = \(3.2 × 60\) = \(192\,C\) Charge \(2e\) liberates one \(Cu^{+2}\) ion No of \(Cu^{+2}\) ion liberate by \(192\,C\) \( = \frac{{192}}{{2e}} = \frac{{192}}{{2 \times 1.6 \times {{10}^{ - 19}}}} = 6 \times {10^{20}}\)
