The specific conductance in ohm- cm- of four electrolytes P, Q, R and S are given in brackets:
`{:(P(5.0 xx 10^(-5)),Q(7.0 xx 10^(-8))) ,(R(1.0 xx 10^(-10)),S(9.2xx 10^(-3))) :}`
The one that offer s highest resistance to the passage of electric current is:

Options

A. P
B. S
C. R
D. Q

Answer

To determine which electrolyte offers the highest resistance to the passage of electric current, we need to calculate the resistance of each electrolyte based on their specific conductance values. The relationship between specific conductance (κ) and resistance (R) is given by the formula:

\[ R = \frac{1}{\kappa} \]

Where:
- R = resistance in ohms (Ω)
- κ = specific conductance in ohm⁻ cm⁻

Given specific conductance values:
- P: \( 5.0 \times 10^{-5} \) ohm⁻ cm⁻
- Q: \( 7.0 \times 10^{-8} \) ohm⁻ cm⁻
- R: \( 1.0 \times 10^{-10} \) ohm⁻ cm⁻
- S: \( 9.2 \times 10^{-3} \) ohm⁻ cm⁻

### Step-by-step Calculation:

1. **Calculate Resistance for P:**
\[ R_P = \frac{1}{5.0 \times 10^{-5}} = 2.0 \times 10^{5} \, \Omega \]

2. **Calculate Resistance for Q:**
\[ R_Q = \frac{1}{7.0 \times 10^{-8}} = 1.4286 \times 10^{7} \, \Omega \]

3. **Calculate Resistance for R:**
\[ R_R = \frac{1}{1.0 \times 10^{-10}} = 1.0 \times 10^{10} \, \Omega \]

4. **Calculate Resistance for S:**
\[ R_S = \frac{1}{9.2 \times 10^{-3}} = 108.7 \, \Omega \]

### Summary of Resistance Values:
- \( R_P = 2.0 \times 10^{5} \, \Omega \)
- \( R_Q = 1.4286 \times 10^{7} \, \Omega \)
- \( R_R = 1.0 \times 10^{10} \, \Omega \)
- \( R_S = 108.7 \, \Omega \)

### Conclusion:
Among the calculated resistances, \( R_R \) has the highest value of \( 1.0 \times 10^{10} \, \Omega \). Therefore, the electrolyte that offers the highest resistance to the passage of electric current is:

**Answer: R**