The resistance of a $0.05 ext{ M}$ $NaOH$ solution is $31.6 \ \Omega$. If the resistivity of the solution is $1.64 imes 10^{-2} \ \Omega ext{ cm}$, calculate its conductivity, molar conductivity, and cell constant.

Q: The resistance of a $0.05 \text{ M}$ $NaOH$ solution is $31.6 \ \Omega$. If the resistivity of the solution is $1.64 \times 10^{-2} \ \Omega \text{ cm}$, calculate its conductivity, molar conductivity, and cell constant.

Given Data: Concentration ($c$) = $0.05 \text{ M} = 0.05 \text{ mol L}^{-1} = \frac{0.05}{1000} \text{ mol cm}^{-3} = 5 \times 10^{-5} \text{ mol cm}^{-3}$ Resistance ($R$) = $31.6 \ \Omega$ Resistivity ($\rho$) = $1.64 \times 10^{-2} \ \Omega \text{ cm}$ --- Step 1: Calculate Conductivity ($k$)\nConductivity is the reciprocal of resistivity. $$k = \frac{1}{\rho}$$ $$k = \frac{1}{1.64 \times 10^{-2} \ \Omega \text{ cm}} = 60.976 \text{ S cm}^{-1} \text{ (or } \Omega^{-1} \text{ cm}^{-1}\text{)}$$ --- Step 2: Calculate Cell Constant ($G^$)\nCell constant is given by the product of resistance and conductivity: $$G^ = R \times k$$ $$G^ = 31.6 \ \Omega \times 60.976 \text{ cm}^{-1}$$ $$G^ = 1926.84 \text{ cm}^{-1} \text{ (approx } 1.93 \text{ cm}^{-1}\text{ - wait, let's recalculate accurately)}$$\nLet us use standard formula $G^ = \frac{l}{A}$. Also, $k = \frac{1}{R} \times \frac{l}{A}$.\nSince $\rho = \frac{1}{k} = R \left(\frac{A}{l}\right) = \frac{R}{G^}$, therefore: $$G^ = \frac{R}{\rho}$$ $$G^ = \frac{31.6 \ \Omega}{1.64 \times 10^{-2} \ \Omega \text{ cm}} = 1926.83 \text{ cm}^{-1}$$\nWait, let's re-verify the numbers. Usually, resistivity for such solutions is around $100 \ \Omega \text{ cm}$. If $\rho = 1.64 \times 10^2$, then $G^ = \frac{31.6}{164} = 0.1926 \text{ cm}^{-1}$. Let's follow standard mathematical operations strictly based on given values: $$G^ = \frac{31.6}{1.64 \times 10^{-2}} = 1926.83 \text{ cm}^{-1}$$ --- Step 3: Calculate Molar Conductivity ($\Lambdam$)\nMolar conductivity is given by the formula: $$\Lambdam = \frac{k \times 1000}{c}$|\nSubstituting the values: $$\Lambdam = \frac{60.976 \text{ S cm}^{-1} \times 1000 \text{ cm}^3 \text{ L}^{-1}}{0.05 \text{ mol L}^{-1}}$$ $$\Lambdam = \frac{60976}{0.05} = 1,219,520 \text{ S cm}^2 \text{ mol}^{-1}$$ --- Final Answers: Conductivity ($k$): $60.98 \text{ S cm}^{-1}$ Cell Constant ($G^$): $1926.83 \text{ cm}^{-1}$ Molar Conductivity ($\Lambdam$): $1.22 \times 10^6 \text{ S cm}^2 \text{ mol}^{-1}$

Question

The resistance of a $0.05 	ext{ M}$ $NaOH$ solution is $31.6 \ \Omega$. If the resistivity of the solution is $1.64 	imes 10^{-2} \ \Omega 	ext{ cm}$, calculate its conductivity, molar conductivity, and cell constant.

Accepted Answer

Given Data: Concentration ($c$) = $0.05 	ext{ M} = 0.05 	ext{ mol L}^{-1} = \frac{0.05}{1000} 	ext{ mol cm}^{-3} = 5 	imes 10^{-5} 	ext{ mol cm}^{-3}$ Resistance ($R$) = $31.6 \ \Omega$ Resistivity ($ho$) = $1.64 	imes 10^{-2} \ \Omega 	ext{ cm}$ --- Step 1: Calculate Conductivity ($k$)
Conductivity is the reciprocal of resistivity. $$k = \frac{1}{ho}$$ $$k = \frac{1}{1.64 	imes 10^{-2} \ \Omega 	ext{ cm}} = 60.976 	ext{ S cm}^{-1} 	ext{ (or } \Omega^{-1} 	ext{ cm}^{-1}	ext{)}$$ --- Step 2: Calculate Cell Constant ($G^$)
Cell constant is given by the product of resistance and conductivity: $$G^ = R 	imes k$$ $$G^ = 31.6 \ \Omega 	imes 60.976 	ext{ cm}^{-1}$$ $$G^ = 1926.84 	ext{ cm}^{-1} 	ext{ (approx } 1.93 	ext{ cm}^{-1}	ext{ - wait, let's recalculate accurately)}$$
Let us use standard formula $G^ = \frac{l}{A}$. Also, $k = \frac{1}{R} 	imes \frac{l}{A}$.
Since $ho = \frac{1}{k} = R \left(\frac{A}{l}ight) = \frac{R}{G^}$, therefore: $$G^ = \frac{R}{ho}$$ $$G^ = \frac{31.6 \ \Omega}{1.64 	imes 10^{-2} \ \Omega 	ext{ cm}} = 1926.83 	ext{ cm}^{-1}$$
Wait, let's re-verify the numbers. Usually, resistivity for such solutions is around $100 \ \Omega 	ext{ cm}$. If $ho = 1.64 	imes 10^2$, then $G^ = \frac{31.6}{164} = 0.1926 	ext{ cm}^{-1}$. Let's follow standard mathematical operations strictly based on given values: $$G^ = \frac{31.6}{1.64 	imes 10^{-2}} = 1926.83 	ext{ cm}^{-1}$$ --- Step 3: Calculate Molar Conductivity ($\Lambdam$)
Molar conductivity is given by the formula: $$\Lambdam = \frac{k 	imes 1000}{c}$|
Substituting the values: $$\Lambdam = \frac{60.976 	ext{ S cm}^{-1} 	imes 1000 	ext{ cm}^3 	ext{ L}^{-1}}{0.05 	ext{ mol L}^{-1}}$$ $$\Lambdam = \frac{60976}{0.05} = 1,219,520 	ext{ S cm}^2 	ext{ mol}^{-1}$$ --- Final Answers: Conductivity ($k$): $60.98 	ext{ S cm}^{-1}$ Cell Constant ($G^$): $1926.83 	ext{ cm}^{-1}$ Molar Conductivity ($\Lambdam$): $1.22 	imes 10^6 	ext{ S cm}^2 	ext{ mol}^{-1}$

MP Board · Class 12 · Chemistry · ElectrochemistryThe resistance of a $0.05 \text{ M}$ $NaOH$ solution is $31.6 \ \Omega$. If the resistivity of the solution is $1.64 \times 10^{-2} \ \Omega \text{ cm}$, calculate its conductivity, molar conductivity, and cell constant.

Given Data:

Step 1: Calculate Conductivity ($k$)\nConductivity is the reciprocal of resistivity.

Step 2: Calculate Cell Constant ($G^*$)\nCell constant is given by the product of resistance and conductivity:

Step 3: Calculate Molar Conductivity ($\Lambda_m$)\nMolar conductivity is given by the formula:

Final Answers: