A 0·05 M NH₄OH solution offers the resistance of 30·8 ohms to a conductivity cell at 298 K. If the…
(a)[3]
A $0\cdot05\text{ M } \text{NH}_4\text{OH}$ solution offers the resistance of $30\cdot8\text{ ohms}$ to a conductivity cell at $298\text{ K}$. If the cell constant is $0\cdot343\text{ cm}^{-1}$ and molar conductance of $\text{NH}_4\text{OH}$ at infinite dilution is $471\cdot4\text{ S cm}^2\text{mol}^{-1}$, calculate the following:
(1) Specific conductance
(2) Molar conductance
(3) Degree of dissociation
(b)[2]
In the diagram of the electrolytic cell given below, A, B and C are connected in series having electrolytes of $\text{ZnSO}_4$, $\text{AgNO}_3$ and $\text{CuSO}_4$ respectively. A steady current of $1\cdot5\text{ A}$ was passed until $1\cdot45\text{ g}$ of $\text{Ag}$ was deposited at the cathode of cell B.
(Atomic mass of $\text{Ag} = 108$, $\text{Cu} = 63\cdot5$, $\text{Zn} = 65\cdot3$)
Answer the following questions.
(1) How long did the current flow?
(2) What weight of Cu and Zn was deposited at cathode?
(1) Charge: $Q = \dfrac{1.45}{108}\times96500 = 1295.6\ \mathrm{C}$ (1 mol of electrons per mol of Ag), so $t = \dfrac{Q}{I} = \dfrac{1295.6}{1.5} = 863.7\ \mathrm{s}$
(2) The same charge flows through all cells; Cu and Zn each need 2 F per mole:
$w_{Cu} = \dfrac{1295.6\times63.5}{2\times96500} = 0.426\ \mathrm{g}$
$w_{Zn} = \dfrac{1295.6\times65.3}{2\times96500} = 0.438\ \mathrm{g}$
