Question

Resistance of Solution-I is 45 Ω and that of Solution-ll is 90 Ω. Both solutions are being analysed in the same conductivity vessel. Now 250 mL of solution -I is mixed with 250 mL of Solution-II. If the resistance of the mixture of p x 109, find p + q, assuming there is no change in '$\alpha$' upon mixing ($\alpha$ is degree of dissociation). (p and q are single digit integer).

Answer 1

7

Answer 2

To solve this problem, we need to understand the relationship between resistance, conductance, conductivity, and cell constant. We also need to consider how mixing solutions affects their overall conductivity.

Formulas Used:

1. Conductance (G): $G = \frac{1}{R}$, where R is resistance.
2. Conductivity ($\kappa$): $\kappa = G \times G^* = \frac{1}{R} \times G^*$, where $G^*$ is the cell constant.

Let $R_1$ be the resistance of Solution-I and $R_2$ be the resistance of Solution-II. Given: $R_1 = 45 \, \Omega$ $R_2 = 90 \, \Omega$

Since both solutions are analyzed in the same conductivity vessel, the cell constant ($G^*$) is the same for both.

For Solution-I: The conductivity $\kappa_1$ is given by: $\kappa_1 = \frac{1}{R_1} \times G^* = \frac{1}{45} \times G^*$

For Solution-II: The conductivity $\kappa_2$ is given by: $\kappa_2 = \frac{1}{R_2} \times G^* = \frac{1}{90} \times G^*$

When 250 mL of Solution-I is mixed with 250 mL of Solution-II, the total volume becomes $250 \, \text{mL} + 250 \, \text{mL} = 500 \, \text{mL}$. Since equal volumes are mixed, the concentration of each solution in the mixture becomes half of its original concentration. The problem states that there is no change in the degree of dissociation ($\alpha$) upon mixing. This implies that the molar conductivity of the ions remains constant, and thus the conductivity contributed by each solution is directly proportional to its concentration in the mixture.

Therefore, the contribution of Solution-I to the mixture's conductivity will be $\frac{\kappa_1}{2}$, and the contribution of Solution-II will be $\frac{\kappa_2}{2}$.

The conductivity of the mixture ($\kappa_{mix}$) is the sum of the conductivities contributed by each component: $\kappa_{mix} = \frac{\kappa_1}{2} + \frac{\kappa_2}{2}$ $\kappa_{mix} = \frac{1}{2} \left( \frac{G^*}{45} + \frac{G^*}{90} \right)$ $\kappa_{mix} = \frac{G^*}{2} \left( \frac{2}{90} + \frac{1}{90} \right)$ $\kappa_{mix} = \frac{G^*}{2} \left( \frac{3}{90} \right)$ $\kappa_{mix} = \frac{G^*}{2} \left( \frac{1}{30} \right)$ $\kappa_{mix} = \frac{G^*}{60}$

Let $R_{mix}$ be the resistance of the mixture. Using the formula for conductivity: $\kappa_{mix} = \frac{1}{R_{mix}} \times G^*$ Substitute the calculated $\kappa_{mix}$: $\frac{G^*}{60} = \frac{1}{R_{mix}} \times G^*$ $\frac{1}{60} = \frac{1}{R_{mix}}$ $R_{mix} = 60 \, \Omega$

The problem states that the resistance of the mixture is $p \times 10^q$, where $p$ and $q$ are single-digit integers. We have $R_{mix} = 60 \, \Omega$. So, $60 = p \times 10^q$.

To find single-digit integers $p$ and $q$: If $q=0$, then $p \times 10^0 = p = 60$, which is not a single-digit integer. If $q=1$, then $p \times 10^1 = 60 \implies p \times 10 = 60 \implies p = 6$. Here, $p=6$ is a single-digit integer, and $q=1$ is a single-digit integer. This satisfies the condition.

Therefore, $p=6$ and $q=1$. We need to find $p+q$. $p+q = 6+1 = 7$.

The final answer is $\boxed{7}$.

Explanation of the solution:

1. Calculate initial conductivities: $\kappa_1 = G^*/45$ and $\kappa_2 = G^*/90$.
2. Determine mixture conductivity: Since equal volumes are mixed and $\alpha$ is constant, $\kappa_{mix} = (\kappa_1 + \kappa_2)/2$.
3. Substitute initial conductivities: $\kappa_{mix} = (G^*/45 + G^*/90)/2 = (2G^*/90 + G^*/90)/2 = (3G^*/90)/2 = G^*/60$.
4. Calculate mixture resistance: $R_{mix} = G^*/\kappa_{mix} = G^*/(G^*/60) = 60 \, \Omega$.
5. Express resistance in given form: $60 = p \times 10^q$. For $p, q$ to be single-digit integers, $p=6$ and $q=1$.
6. Calculate $p+q$: $6+1=7$.