Question

The specific conductance (conductivity) of an M/1000 solution of an acid HA is 6 x 10-5 S cm at
298K. am of Hi and A- at 298K are 350 and 50 S cm2 mol-1. The % dissociation of the acid in its M/
1000 solution is :​

Answer 1

Answer:

molar conductance=1000k/m=6×10^-5×10^3/10^-3=60

molar conductance at infinite dilution=350+50=400 here

degree of dissociation dilution=molar conductance/molar conductance at infinite dilution=60/400=15%

hope it helps.

please mark me as the brainliest.

Answer 2

Correct Question

The specific conductance (conductivity) of an M/1000 solution of an acid HA is 6 x 10-5 $ohm^{-1} \ cm^{-1}$ at 298 K. The molar conductance at infinite dilution of  of $H^{+}$ ion  and $A^{-}$ ion at 298 K are 350 and 50 $ohm^{-1} \ cm^{2} \ mol^{-1}$. The % dissociation of the acid in its M/1000 solution is :​

Answer:

The % dissociation of the acid in its M/1000 solution is 15%.

Explanation:

Given:

The specific conductance of an M/1000 solution of an acid HA is $6\times10^{-5}$ $ohm^{-1} \ cm^{-1}$ at 298 K.

The molar conductance at infinite dilution of  of $H^{+}$ ion  and $A^{-}$ ion at 298 K are 350 and 50 $ohm^{-1} \ cm^{2} \ mol^{-1}$.

To Find:

The % dissociation of the acid in its M/1000 solution.

Solution:

As given,the specific conductance of an M/1000 solution of an acid HA is $6\times10^{-5}$  $ohm^{-1} \ cm^{-1}$ at 298 K.

The specific conductance of,K =$6\times10^{-5}$ $ohm^{-1} \ cm^{-1}$.

For M/1000 solution of an acid HA, the value of  C=0.001

We know  formula

$Molar\ Conductance\ of\ HA\ , \Lambda _{m} ^{c} =\frac{K\times1000}{C}$

                                                    $=\frac{6\times10^{-5}\times1000}{0.001}$

                                                    $=60 \ ohm^{-1} \ cm^{2} \ mol^{-1}$

As  given,The molar conductance at infinite dilution of  of $H^{+}$ ion  and $A^{-}$ ion at 298 K are 350 and 50 $ohm^{-1} \ cm^{2} \ mol^{-1}$.

The molar conductance at infinite dilution of $H^{+}$ ion, $\Lambda ( H^{+} )= 350\ ohm^{-1} \ cm^{2} \ mol^{-1}$

The molar conductance at infinite dilution of  $A^{-}$ion,$\Lambda ( A^{-} )= 50\ ohm^{-1} \ cm^{2} \ mol^{-1}$

$\Lambda_{m} (HA) \infty=\Lambda(H^{+} )+\Lambda(A^{-} )$

                 $=350+50= 400 \ ohm^{-1} \ cm^{2} \ mol^{-1}$

We know  formula

Degree of dissociation of the acid,$\alpha =\frac{\Lambda _{m} ^{c} }{\Lambda_{m} (HA) \infty}$

                                                             $=\frac{60}{400} =0.15$

% of Degree of dissociation of the acid,$\alpha$ $=0.15 \times 100$

                                                                     $=15$

Thus, the % dissociation of the acid in its M/1000 solution is 15%.