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Standard XII

Chemistry

Conductivity of Ionic Solutions

The conductiv...

Question

The conductivity of sodium chloride at $298 K$ has been determined at different concentrations and the results are given below:

Concentration/M $0.001$ $0.010$ $0.020$ $0.050$ $0.100$
$10^{2} k / S m^{- 1}$ $1.237$ $11.85$ $23.15$ $55.53$ $106.74$

Calculate $\land_{m}$ for all concentrations and draw a plot between $\land_{m}$ and $c^{1 / 2}$ . Find the value of $\land_{m}^{0}$ .

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Solution

The plot is shown below.
The equation of plot is $Λ_{m} = - 51286 C^{3} + 9416. C^{2} - 602.5 C + 124.1$
When $C = 0$ M
$Λ_{m}^{0} = - 51286 (0)^{3} + 9416. (0)^{2} - 602.5 (0) + 124.1$
$Λ_{m}^{0} = 124.1 S c m^{2} / m o l$
Note: For calculating $Λ_{m}$ , we have used the formula $Λ_{m} = \frac{1000 κ}{C}$

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The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below :

$\begin{matrix} \frac{C o n c e n t r a t i o n}{M} & 0.001 & 0.010 & 0.020 & 0.050 & 0.100 10^{- 2} \times \frac{k}{S m^{- 1}} & 1.237 & 11.85 & 23.15 & 55.53 & 106.74 \end{matrix}$

Calculate $λ_{m}$ for all concentrations and draw a plot between $λ_{m}$ and $C^{\frac{1}{2}}$ Find the value of $λ_{m}^{\circ}$ .

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M 0.001 0.010 0.020 0.050 0.100

10² × κ/S m⁻¹ 1.237 11.85 23.15 55.53 106.74

Calculate for all concentrations and draw a plot between and c½. Find the value of.

Expt. No.	Initial Concentration		Initial rate
Expt. No.	$[N O]$	$[C l_{2}]$	Initial rate
1.	0.010	0.010	$1.2 \times 10^{- 4}$
2.	0.010	0.020	$2.4 \times 10^{- 4}$
3.	0.020	0.020	$9.6 \times 10^{- 4}$

For the reaction,

2 N O + {C l}_{2} ⟶ 2 N O C l

, at

300 K

, following data are obtained:
Write rate law and order of the reaction? Also, calculate the specific rate constant.

Q. The table below shows recorded concentration data for the following chemical reaction:

C a C l_{2} (a q) + 2 A g N O_{3} (a q) \to C a (N O_{3})_{2} (a q) + 2 A g C l (s)

Use the data to determine the exponents

x

and

y

in the rate law:

r a t e = k [C a C l_{2}]^{x} [A g N O_{3}]^{y}

$[C a C l_{2}] M$	$[A g N O_{3}] M$	Rate M/s
$0.050$	$0.050$	$3.33 \times 10^{- 3}$
$0.100$	$0.050$	$6.66 \times 10^{- 3}$
$0.100$	$0.100$	$13.32 \times 10^{- 3}$

Q. The electrochemical cell shown below is a concentration cell.

M | M^{2 +}

(Saturated solution of a sparingly soluble salt,

M X_{2}) | | M^{2 +} (0.001 m o l d m^{- 3}) M .

The emf of the cell depends on the difference in concentration of

M^{2 +}

ions at the two electrodes. The emf of the cell at 298K is 0.059

The value of

Δ G (k J m o l^{- 1})

for the given cell is: (take

1 F = 96500 C m o l^{- 1})

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Concentration/M	$0.001$	$0.010$	$0.020$	$0.050$	$0.100$
$10^{2} k / S m^{- 1}$	$1.237$	$11.85$	$23.15$	$55.53$	$106.74$

The plot is shown below.The equation of plot is Λm=−51286C3+9416.C2−602.5C+124.1When C=0 MΛ0m=−51286(0)3+9416.(0)2−602.5(0)+124.1Λ0m=124.1Scm2/molNote: For calculating Λm, we have used the formula Λm=1000κC