The molar conductivities $$\Lambda _{NaOAc}^o$$ and $$\Lambda _{HCl}^o$$ at infinite dilution in water at $${25^ \circ }C$$ are $$91.0$$ and $$426.2\,S\,c{m^2}/mol$$ respectively. To calculate $$\Lambda _{HOAc}^o,$$ the additional value required is
A.
$$\Lambda _{NaOH}^o$$
B.
$$\Lambda _{NaCl}^o$$
C.
$$\Lambda _{{H_2}O}^o$$
D.
$$\Lambda _{KCl}^o$$
Answer :
$$\Lambda _{NaCl}^o$$
Solution :
$$\Lambda _{C{H_3}COOH}^o$$ is given by the following equation
$$\Lambda _{C{H_3}COOH}^o = \left( {\Lambda _{C{H_3}COONa}^o + \Lambda _{HCl}^o} \right) - \left( {\Lambda _{NaCl}^o} \right)$$
Hence $${\Lambda _{NaCl}^o}$$ is required.
Releted MCQ Question on Physical Chemistry >> Electrochemistry
Releted Question 1
The standard reduction potentials at $$298 K$$ for the following half reactions are given against each
$$\eqalign{
& Z{n^{2 + }}\left( {aq} \right) + 2e \rightleftharpoons Zn\left( s \right)\,\,\,\,\,\,\,\,\, - 0.762 \cr
& C{r^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons Cr\left( s \right)\,\,\,\,\,\,\,\,\, - 0.740 \cr
& 2{H^ + }\left( {aq} \right) + 2e \rightleftharpoons {H_2}\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.000 \cr
& F{e^{3 + }}\left( {aq} \right) + 2e \rightleftharpoons F{e^{2 + }}\left( {aq} \right)\,\,\,\,\,\,\,\,0.770 \cr} $$
which is the strongest reducing agent ?
A solution containing one mole per litre of each $$Cu{\left( {N{O_3}} \right)_2};AgN{O_3};H{g_2}{\left( {N{O_3}} \right)_2};$$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are :
$$\eqalign{
& Ag/A{g^ + } = + 0.80,\,\,2Hg/H{g_2}^{ + + } = + 0.79 \cr
& Cu/C{u^{ + + }} = + 0.34,\,Mg/M{g^{ + + }} = - 2.37 \cr} $$
With increasing voltage, the sequence of deposition of metals on the cathode will be :