Questions
What do you mean by equivalent Conductivity?
- It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution.
- It is defined as the conducting power of all the ions produced by dissolving ten gram equivalent of an electrolyte in solution.
- It is defined as the conducting power of all the ions produced by dissolving hundred gram equivalent of an electrolyte in solution.
- It is defined as the conducting power of all the ions produced by dissolving thousand gram equivalent of an electrolyte in solution.
The resistance of $1\ N$ solution of $CH _{3}COOH$ is $250\ ohm$ when measured in a cell of cell constant $1.15\ cm^{-1}$. The equivalent conductance will be:
- $4.6\ ohm^{-1} cm^{2} eq^{-1}$
- $9.2\ ohm^{-1} cm^{2} eq^{-1}$
- $18.4\ ohm^{-1} cm^{2} eq^{-1}$
- $0.023\ ohm^{-1} cm^{2} eq^{-1}$
The specific conductance of a $0.01\ M$ solution of $KCl$ is $0.0014\ ohm^{-1} cm^{-1}$ at $25^{\circ}C$. Its equivalent conductance is____________.
- $14$
- $140$
- $1.4$
- $0.14$
The resistance of a N/10 KCI solution is 245$\Omega $. Calculate the equivalent conductance of the solution if the electrodes in the cell are 4cm apart and each having an area of 7.0sq,cm.
- $23.32S{ cm }^{ 2 }{ eq }^{ -1 }$
- $23.23S{ cm }^{ 2 }{ eq }^{ -1 }$
- $2.332S{ cm }^{ 2 }{ eq }^{ -1 }$
- none of these
The equivalent conductance of a weak monobasic acid at infinite dilution is $100cm^3$ $eq^{-1}$ and that of its $0.01$M solution is $5cm^2$ $eq^{-1}$ at $25^o$C. The dissociation constant $K _a$ of the acid is:
- $2.5\times 10^{-4}$
- $5\times 10^{-4}$
- $1.25\times 10^{-5}$
- $2.5\times 10^{-5}$
Equivalent conductance and molar conductance of $Fe _2(SO _4) _3$ are related?
- $\bigwedge _e=\bigwedge _m$
- $\bigwedge _{eq} = \dfrac{\bigwedge _m}{3}$
- $\bigwedge _{eq} = 3 \bigwedge _m$
- $\bigwedge _{eq} = \dfrac{\bigwedge _m}{6}$
Molar ionic conductance of ${Ca}^{+2}$ is $x$ $S{m}^{2}$ ${mole}^{-1}$. Equivalent conductance of calcium phosphate is ____ $S{m}^{2}g$ ${eq}^{-1}$
- $x+y$
- $(3x+2y)$
- $6(3x+2y)$
- $\cfrac{3x+2y}{6}$
Which of the following has least conductivity in aqueous solution?
- $CO(NH _3) _4Cl _3$
- $Co(NH _3) _3Cl _3$
- $Co(NH _3) _5Cl _3$
- $Co(NH _3) _6Cl _3$
The equivalent conductance of $CH _3COONa, \ HCl$ and $NaCl$ at infinite dilution are $91, 426$ and $126 \ S \ cm^3 \ eq^{-1}$ respectively at $25^oC$. The equivalent conductance of $1 \ M \ CH _3COOH$ solution is $19.55 \ S \ cm^2 \ eq^{-1}$. The pH of the solution is:
- $5.3$
- $4.3$
- $2.3$
- $1.3$
The electrolytic conductance is a direct measure of:
- resistance
- potential
- concentration
- dissociation
Equivalent conductance of $BaCl _2, H _2SO _4$ and $HCl$ are $x _1, x _2$ and $x _3 S cm^2 equiv^{-1}$ at infinite dilution.If specific conductance of saturated $BaSO _4$ solution is of $y S cm^1$ then $K _{sp}$ of $BaSO _4$ is:
- $\frac{10^3y}{2(x _1 + x _2 - 2x _3)}$
- $\frac{10^6y^2}{4(x _1 + x _2 - 2x _3)^2}$
- $\frac{10^6y^2}{2(x _1 + x _2 - x _3)^2}$
- $\frac{x _1 + x _2 - 2x _3}{10^6y^2}$
For $HCl$ solution at ${25}^{o}C$ equivalent conductance at infinite dilution is $425 \ {ohm}^{-1}{cm}^{2}{equiv}^{-1}$. The specific conductance of a solution of $HCl$ is $3.825$ ${ohm}^{-1}{cm}^{-1}$. If the apparent degree of dissociation is $90$% the normality of the solution is :
- $0.90N$
- $1.0N$
- $10\ N$
- $1.2N$
The equivalent conductivity of $0.1 N \ CHNCH _{3}COOH$ at $25^{0}C$ is 80 and at infinite dilution it is 400, the degree of dissociation of $CH _{3}COOH$ is :
- 1
- 0.2
- 0.1
- 0.5
In infinite dilusions, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ are $127$ and $76 ohm^{-1} , cm^{-1} , eqvt^{-1}$. The equivalent conductivity of $BaCl _2$ at indefinite dilution is?
- $101.5$
- $203.5$
- $139.5$
- $279.5$
${\text{N}}{{\text{a}} _{\text{3}}}{\text{Al}}{{\text{F}} _{\text{6}}},,$ is added to $,{\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}$
- Improve the electrical conductivity of the cell
- Increases rate of production
- Increases the melting point
- Decrease the electrical conductivity
The equivalent conductance of $0.02$ M acctic acid $1.62.*{10^{ - 3}}$. Degree of ironisation $'a'$ of $C{H _3}COOH$ is:
$({x _H} = 349.83oh{m^{ - 1}}and\lambda C{H _3}CO{O^ - } = 40.89ohm{s^{ - 1}}$
- $0.01$
- $0.02$
- $0.03$
- $0.04$
Equivalent conductivity of $BaCl _2,H _2SO _4$ and HCI, are $x _1,x _2$ and $x _3scm^{-1}eq^{-1}$ at infinite dilution. If conductivity of saturated $BaSO _4$ solution is x $Scm^{-1}$, then $K _{sp}$ of $BaSO _4$ is
- $\dfrac {500x} {(x _1+x _2-2x _3)}$
- $\dfrac {10^6x^2} {(x _1+x _2-2x _3)^3}$
- $\dfrac {2.5\times10^5 x^2} {x _1-2x _2-x _3)^2}$
- $\dfrac {0.25 x^2} {x _1 + x _2-x _3)^2}$
Molar conductance of $C{a^{2 + }}$ and $C{l^ - }$ are $120,c{m^2},$ $mo{l^{ - 1}}$ and $77,S,c{m^2},mo{l^{ - 1}}$ respectively. What is the equivalent conductance of $CaC{l _2}$ ?
- $98.5\,Sc{m^2}e{q^{ - 1}}$
- $137\,Sc{m^2}e{q^{ - 1}}$
- $197\,Sc{m^2}e{q^{ - 1}}$
- $247\,Sc{m^2}e{q^{ - 1}}$
At infinite dilution equivalent conductance of ${B^{ + 2}}$ & CI ions are 127 & 76$oh{m^{ - 1}}$ $c{m^{ - 1}}$` $e{q^{ - 1}}$ respectively. Equivalent conductance of $BaC{I _2}$ at infinite diluition is :
- 139.5
- 101.5
- 203
- 279
Equivalent constant of standard $BaSO _{4}$ is $400ohm^{-1}\ cm^{2}$ equiv$^{-1}$ and specific conduction is $8\times 10^{-5}\ ohm^{-1}\ cn^{-1}$. Hence $K _{SP}$ of $BaSO _{4}$ is
- $4\times 10^{-8}M^{2}$
- $1\times 10^{-8}M^{2}$
- $2\times 10^{-4}M^{2}$
- $1\times 10^{-4}M^{2}$
Which of the following is correct regarding current carrying ions in the solution of $C _{2}H _{5}COOH$ upon dilution?
- The number of ions in $1 cm^{3}$, as well as in total volume increases.
- The number of ions in $1 cm^{3}$ decreases, whereas that in the total volume remains constant.
- The number of ions in $1 cm^{3}$ decreases, but that in the total volume increases.
- The number of ions in $1 cm^{3}$, as well as in total volume decreases.
The resistance of $N/2$ solution of an electrolyte in a cell was found to be $45$ ohm. The equivalent conductivity of a solution, if the electrodes in the cell are $2.2$ cm apart and have an area of $3.8 cm^2$ will be:
- $52.72\ S cm^2 eq^{-1}$
- $22.57\ S cm^2 eq^{-1}$
- $27.52\ S cm^2 eq^{-1}$
- $25.72\ S cm^2 eq^{-1}$
The resistance of 0.1 N solution of a salt is found to be $2.5\times10^{3}$. The equivalent conductance of the solution is: (cell constant=1.15 $cm^{-1}$)
- 3.6
- 4.6
- 5.6
- 6.6
At 291 K, the equivalent conductivities at infinite dilution of $NH _4Cl$, $NaOH $ and $NaCl $ are $129.8$, $217.5$ and $108.9$ S $cm^2 eq^{-1}$ respectively. The equivalent conductivity at infinite dilution of $NH _4OH$ is:
- $208.4 S cm^2 eq^{-1}$
- $238.4 S cm^2 eq^{-1}$
- $283.4 S cm^2 eq^{-1}$
- None of these
The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is:
- $LiCl > NaCl > KCl$
- $KCl > NaCl > LiCl$
- $NaCl > KCl > LiCl$
- $LiCl > KCl > NaCl$
$\lambda _{eq}$ x Normality = __________
- K x 10$^3$
- K x 10$^4$
- K x 10$^5$
- K x 10$^6$
Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1\ Ohm^{-1} cm^{2}$. If $NaF$ is replaced by $KF$ what is the value of equivalent conductance?
- $90.1\ Ohm^{-1} cm^{2}$
- $11.2\ Ohm^{-1} cm^{2}$
- $0$
- $222.4\ Ohm^{-1} cm^{2}$
The resistance of $N/10$ solution is found to be $2.5\times 10^{3}ohm$. The equivalent conductance of the solution is (cell constant $= 1.25\ cm^{-1})$.
- $2.5\ ohm^{-1} cm^{2} equiv^{-1}$
- $5\ ohm^{-1} cm^{2} equiv^{-1}$
- $2.5\ ohm^{-1} cm^{-2} equiv^{-1}$
- $5\ ohm^{-1} cm^{-2} equiv^{-1}$
What are the units of equivalent conductivity of a solution?
- $mho\ cm^{-1}$
- $ohm\ cm^{-1} g\ equiv^{-1}$
- $mho\ cm^{-2}g\ equiv^{-1}$
- $mho\ cm^{2}g\ equiv^{-1}$
The equivalent conductances of $NaCl$ at concentration $c$ and at infinite dilution are $\lambda _{c}$ and $\lambda _{\infty}$ respectively. The correct relationship between $\lambda _{c}$ and $\lambda _{\infty}$ is given as: (where the constant $b$ is positive).
- $\lambda _{c} = \lambda _{\infty} - b\sqrt {c}$
- $\lambda _{c} = \lambda _{\infty} + b\sqrt {c}$
- $\lambda _{c} = \lambda _{\infty} + bc$
- $\lambda _{c} = \lambda _{\infty} - bc$
The equivalent conductances at infinite dilution of $HCl$ and $NaCl$ are $426.15$ and $126.15\ mho\ cm^{2}g\ eq^{-1}$ respectively. If can be said that the mobility of:
- $H^{+}$ ions is much more than that of $Cl^{-}$ ions
- $Cl^{-}$ ions is much more than that of $H^{+}$ ions
- $H^{+}$ ions is much more than that of $Na^{+}$ ions
- $Na^{+}$ ions is much more than that of $H^{+}$ ions
The resistance of $0.01\ N$ solution at $25^{\circ}$ is $200\ ohm$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.
- $200\ ohm^{-1}cm^{2} eq^{-1}$.
- $300\ ohm^{-1}cm^{2} eq^{-1}$.
- $400\ ohm^{-1}cm^{2} eq^{-1}$.
- $500\ ohm^{-1}cm^{2} eq^{-1}$.
The specific conductance $(K)$ of an electrolyte of $0.1\ N$ concentration is related to equivalent conductance $(\wedge _{e})$ by the following formula.
- $\wedge _{e} = K$
- $\wedge _{e} = 10 K$
- $\wedge _{e} = 100 K$
- $\wedge _{e} = 10000 K$
The specific conductivity of $0.1$ $N$ $KCl$ solution at $20^0$C is $0.0212$ $ohm^{-1} cm^{-1}$. The solution was found to offer resistance of $55$ ohms. Find the cell constant of the conductivity cell.
- $2.25$
- $1.166$
- $1.936$
- none of these
What would be the equivalent conductivity of a cell in which $0.5$ N salt solution offers a resistance of $40$ ohm whose electrodes are $2$ cm apart and $5$ $cm^{2}$ in area?
- $10$ $ohm^{-1}$ $cm^2$ $eq^{-1}$
- $20$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
- $30$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
- $25$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
The equivalent conductivity of $N/10$ solution of acetic acid at $25^o$C is $14.3$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$. What will be the degree of dissociation of acetic acid?
$(\Lambda _{\infty CH _3COOH}=390.71$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$).
- $3.66\%$
- $3.9\%$
- $2.12\%$
- $0.008\%$
Which of the following statements is true?
- When an aqueous solution of NaCl is electrolysed, sodium metal is deposited at cathode
- There is no difference between specific conductivity and molar conductivity
- Silver nitrate solution can be stored in a copper container
- The addition of liquid bromine to iodide solution turns it violet
The ionic conductivity of $B{a^{2 + }}$ and $C{l^ - }$ at infinite dilution are 127 and 76 respectively. The equivalent conductivity of $BaC{l _2}$ at infinite dilution (in $oh{m^ - },c{m^2},e{q^{ - 1}}$) would be:
- 279
- 280
- 139.5
- 102
The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:
- LiCl $>$ NaCl $>$ KCl
- KCl $>$ NaCl $>$ LiCl
- NaCl $>$ KCl $>$ LiCl
- LiCl $>$ KCl $>$ NaCl
The equivalent conductivity conductivity of 1M ${{\text{H}} _{\text{2}}}{\text{S}}{{\text{O}} _{\text{4}}}$ solution would be if specific conductance is ${\text{26}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{S}},{\text{c}}{{\text{m}}^{ - 1}}$.
- $1.3 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{e}}{{\text{q}}^{ - 1}}$
- $1.6 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,$
- $13\,{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
- $1.3 \times {10^3}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Equivalent conductance at infinite dilution for weak electrolyte HF:
- can be determined by measurement of equivalent conductance at infinite dilution for dilute solution of $HCL, \; HBr$ and $HI$
- can be determined by measurement of equivalent conductance at infinite dilution for very dilute $HF$ solutions
- can best be determined from measurements on dilute
- can not be calculated
The conductivity of a saturated solution of $Ba{ SO } _{ 4 }$ is $306\times { 10 }^{ -6 }{ ohm }^{ -1 }{ cm }^{ -1 }$ and its equivalent conductance is $1.53 \ { ohm }^{ -1 }{ cm }^{ 2 }{ equiv }^{ -1 }$.
- $4\times { 10 }^{ -12 }$
- $2.5\times { 10 }^{ -9 }$
- $2.5\times { 10 }^{ -13 }$
- $4\times { 10 }^{ -6 }$
If the specific resistance of a solution of concentration C g equivalent/litre is R, then its equivalent conductance is:
- $\dfrac{100R}{C}$
- $\dfrac{RC}{1000}$
- $\dfrac{1000}{RC}$
- $\dfrac{C}{1000R}$
The resistance of $0.2\ M$ solution of an electrolyte is $50\ \Omega$.The specific conductance of the solution is $1.3\ S\ m^{-1}$. If the resistance of the $0.4\ M$ solution of the same electrolyte is $260\ \Omega$, its molar conductivity is :
- $62.5\ S\ m^{2} mol^{-1}$
- $6250\ S\ m^{2} mol^{-1}$
- $6.25\ \times10^{-4}S\ m^{2} mol^{-1}$
- $625\times10^{-4}\ S\ m^{2} mol^{-1}$
The equivalent conductivity of monobasic acid at infinite dilution is 348 $ohm^{-1}$ $cm^2$ $eq^{-1}$. If the resistivity of the solution containing 15 g acid (molar mass 49) in 1 litre is 18.5 ohm cm, what is the degree of dissociation of acid?
- 45.9%
- 40.2%
- 60.4%
- 50.7%
The conductivities at infinite dilution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{Cl,NaOH}}$ and $\text{NaCl}$ are 130, 218, 120 ${\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$. If equivalent conductance of N/100 solution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ is 10, then degree of dissociation of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ at this dilution is:
- 0.005
- 0.043
- 0.01
- 0.02