At 25^° C , the molar conductance at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl_2 are 248 × 10^-4, 126 × 10^-4 and 280 × 10^-4 Sm^2 ~mol^-1 respectively. Δ_m^0 Ba(OH)_2 in Sm^2 ~mol^-1 is

Correct answer: (B) 524 × 10^-4. Δ_Na^+^°+Δ_OH^-^°=248 × 10^-4 Sm^2 ~mol^-1 Δ_Na^+^°+Δ_Cl^-^°=126 × 10^-4 Sm^2 ~mol^-1 Δ_Ba^°^2++Δ _2 Cl^-^°=280 × 10^-4 Sm^2 ~mol^-1 NoW , Δ_Ba(OH)_2^°=Δ_BaCl_2^°+2 Δ^° _NaOH^°-2 Δ _NaC 1^° Δ _Ba(OH)_2^°=280 × 10^-4+2 × 248 × 10^-4-2 × 126 × 10^-4 Δ_Ba(OH)_2^°=524 × 10^-4 Sm^2 ~mol^-1

Q40 / 125 · BITSAT 2021 · Chemistry

At $25^{\circ} \mathrm{C}$, the molar conductance at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_{2}$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ respectively. $\Delta_{\mathrm{m}}{ }^{0} \mathrm{Ba}(\mathrm{OH})_{2}$ in $\mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ is

A
$52.4 \times 10^{-4}$
Correct answer
B
$524 \times 10^{-4}$
C
$402 \times 10^{-4}$
D
$262 \times 10^{-4}$

Solution

$\Delta_{\mathrm{Na}^{+}}^{\circ}+\Delta_{\mathrm{OH}^{-}}^{\circ}=248 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\Delta_{\mathrm{Na}^{+}}^{\circ}+\Delta_{\mathrm{Cl}^{-}}^{\circ}=126 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\Delta_{\mathrm{Ba}}^{\circ}{ }^{2+}+\Delta{ }_{2 \mathrm{Cl}^{-}}^{\circ}=280 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ $\mathrm{NoW}$, $\Delta_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=\Delta_{\mathrm{BaCl}_{2}}^{\circ}+2 \Delta{ }^{\circ}{ }_{\mathrm{NaOH}}^{\circ}-2 \Delta{ }_{\mathrm{NaC} 1}^{\circ}$ $\Delta{ }_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=280 \times 10^{-4}+2 \times 248 \times 10^{-4}-2 \times 126 \times 10^{-4}$ $\Delta_{\mathrm{Ba}(\mathrm{OH})_{2}}^{\circ}=524 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$

Hence, the correct answer is option: 'B'.

Solution

More Electrochemistry Questions