If the values of wedge-infty- of NH-4-Cl- NaOH and NaCl are 130- 217 and 109 ohm-1- cdot cm-2-cdot equiv-1- respectively- the wedge-infty- of NH-4-OH in ohm-1- cdot cm-2- cdot equiv-1- is-45622238198

Question

Toppr Admin · Accepted Answer

-xA0-x2227-x221E-NH4OH-x2227-x221E-NH4Cl-x2227-x221E-NaOH-x2212-x2227-x221E-NaCl-xA0-x2227-x221E-NH4OH-130-217-x2212-109-xA0-x2227-x221E-NH4OH-238ohm-x2212-1-x22C5-cm2-x22C5-equiv-x2212-1

If the values of $\land_{\infty}$ of $N H_{4} C l, N a O H$ and $N a C l$ are $130, 217$ and $109 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ respectively, the $\land_{\infty}$ of $N H_{4} O H$ in $o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ is:
$456$
$22$
$238$
$198$

$\land_{\infty (N H_{4} O H)} = \land_{\infty (N H_{4} C l)} + \land_{\infty (N a O H)} - \land_{\infty (N a C l)}$

$\land_{\infty (N H_{4} O H)} = 130 + 217 - 109$

$\land_{\infty (N H_{4} O H)} = 238 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$

If the values of ∧∞ of NH4Cl,NaOH and NaCl are 130,217 and 109 ohm−1⋅cm2⋅equiv−1 respectively, the ∧∞ of NH4OH in ohm−1⋅cm2⋅equiv−1 is:45622238198

∧∞(NH4OH)=∧∞(NH4Cl)+∧∞(NaOH)−∧∞(NaCl) ∧∞(NH4OH)=130+217−109 ∧∞(NH4OH)=238ohm−1⋅cm2⋅equiv−1

If the values of $\land_{\infty}$ of $N H_{4} C l, N a O H$ and $N a C l$ are $130, 217$ and $109 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ respectively, the $\land_{\infty}$ of $N H_{4} O H$ in $o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ is:
$456$
$22$
$238$
$198$

$\land_{\infty (N H_{4} O H)} = \land_{\infty (N H_{4} C l)} + \land_{\infty (N a O H)} - \land_{\infty (N a C l)}$

$\land_{\infty (N H_{4} O H)} = 130 + 217 - 109$

$\land_{\infty (N H_{4} O H)} = 238 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$