At 291 K the molar conductivities at infinite dilution of NH4Cl,NaOH and NaCI are 129.8, 217.4 and 108.9 Scm2 respectively,. If the molar conductivity of a centinormal solution of NH4OH is 9.33 Scm2, What is the precentage dissociation of NH4OH at this dilution?

Question

At 291 K the molar conductivities at infinite dilution of NH4Cl,NaOH and NaCI are 129.8, 217.4 and 108.9 Scm2 respectively,. If the molar conductivity of a centinormal solution of NH4OH is 9.33 Scm2, What is the precentage dissociation of NH4OH at this dilution? (A) 1.76% (B) 2.28% (C) 3.92% (D) 4.15%

Tardigrade Admin · Accepted Answer

From Kohlrauschs law, Ao for NH4OH= ANH4+o+ACH-o=Ao(NH4Cl)+Ao(NaOH) -Ao(NaCl)=129.8+217.4-108.9 =238.3 S cm2 Degree of dissociation (α )=(Ac/Ao)=(9.33/238.3)=0.0392=3.92%

Q. At 291 K the molar conductivities at infinite dilution of NH4​Cl,NaOH and NaCI are 129.8, 217.4 and 108.9Scm2 respectively,. If the molar conductivity of a centinormal solution of NH4​OH is 9.33Scm2, What is the precentage dissociation of NH4​OH at this dilution?

1.76%

2.28%

3.92%

4.15%

From Kohlrauschs law, Ao for NH4​OH= ANH4+​o​+ACH−o​=Ao(NH4​Cl)+Ao(NaOH) −Ao(NaCl)=129.8+217.4−108.9 =238.3Scm2 Degree of dissociation (α)=AoAc​​=238.39.33​=0.0392=3.92

From Kohlrauschs law, $A^{o}$ for $N H_{4} O H =$ $A_{N H_{4}^{+}}^{o} + A_{C H^{-}}^{o} = A^{o} (N H_{4} Cl) + A^{o} (N a O H)$ $- A^{o} (N a Cl) = 129.8 + 217.4 - 108.9$ $= 238.3 S c m^{2}$ Degree of dissociation $(α) = \frac{A _{c}}{A ^{o}} = \frac{9.33}{238.3} = 0.0392 = 3.92$