A conductivity cell is used to measure the conductance of electrolyte . It makes use of conductivity of water which does not contain any ions. The cell constant of conductivity cell is determined.
If the cell constant is `0.40 cm^(-1)`, the conductivity of `0.051 M NaCl` solution having `R=1850 ohm` is equal to

Options

A. `1.08xx10^(-4)S cm^(-1)`
B. `4.32 xx 10^(-4) S cm^(-1)`
C. `2.16xx10^(-4) S cm^(-1)`
D. `5.04xx10^(-5) S cm^(-1)`

Explanation

To solve the problem, we will use the relationship between conductivity (κ), conductance (G), and the cell constant (G*). Here are the steps to find the conductivity of the 0.051 M NaCl solution:

### Step 1: Understand the relationship between conductance and resistance
Conductance (G) is the reciprocal of resistance (R):
\[ G = \frac{1}{R} \]

### Step 2: Calculate the conductance (G)
Given that the resistance (R) is 1850 ohms, we can calculate the conductance:
\[ G = \frac{1}{1850 \, \Omega} \]
\[ G = 0.00054054 \, S \] (where S is siemens)

### Step 3: Use the cell constant to find conductivity (κ)
The conductivity (κ) is given by the formula:
\[ \kappa = G \times G^* \]
where \( G^* \) is the cell constant. Given that \( G^* = 0.40 \, cm^{-1} \):
\[ \kappa = 0.00054054 \, S \times 0.40 \, cm^{-1} \]

### Step 4: Calculate the conductivity (κ)
Now we can perform the multiplication:
\[ \kappa = 0.00054054 \times 0.40 \]
\[ \kappa = 0.000216216 \, S/cm \]
This can be expressed in scientific notation:
\[ \kappa \approx 2.16 \times 10^{-4} \, S/cm \]

### Final Answer
The conductivity of the 0.051 M NaCl solution is approximately:
\[ \kappa \approx 2.16 \times 10^{-4} \, S/cm \]

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