Plz answer 3.14 

Asked by lovemaan5500 | 13th Aug, 2019, 05:47: PM

Expert Answer:

Given:
 
Molarity = 0.05 M
 
Resistance = 410.5 Ω
 
Conductance 
 
equals 1 over straight R

equals fraction numerator 1 over denominator 410.5 end fraction
 
 =0.0024 / ohm
 
Specific Conductance = Cell constant × conductance
 
Cell constant = 
 
equals fraction numerator 0.00189 over denominator 0.0024 end fraction

equals 0.7758 space cm to the power of negative 1 end exponent
 
Speficific conductance =  Cell constant × conductance
 
Cell constant is same for both the cells.
 
Speficific conductance K (CaCl2) = 
 
equals space 0.7758 space cross times bevelled 1 over 990 space space

equals space space 7.83 cross times 10 to the power of negative 4 end exponent space space ohm to the power of negative 1 end exponent cm to the power of negative 1 end exponent            
 
 
Normality of CaCl2 =fraction numerator No. space of space gram space equivalents over denominator Volume space of space solution space in space straight L end fraction
 
No. of gram equivalents of CaCl2 =
 
 fraction numerator Weight space of space CaCl subscript 2 over denominator Equivalent space weight space end fraction

equals fraction numerator 11 over denominator begin display style bevelled 111 over 2 end style end fraction

equals fraction numerator 11 over denominator 55.5 end fraction

equals 0.1981 space
 
Normality of CaCl2 =
 
equals fraction numerator 0.1981 over denominator 0.5 end fraction

equals 0.3963 space straight N
 
Equivalent conductance Λ=
 
 
 straight capital lambda equals fraction numerator straight K cross times 1000 over denominator straight C end fraction

space space space equals fraction numerator 7.83 cross times 10 to the power of negative 4 end exponent cross times 1000 over denominator begin display style 0.3963 end style end fraction

space space space equals 1.9769 space ohm to the power of negative 1 end exponent cm squared eq to the power of negative 1 end exponent
 
Molar conductance of CaCl2 =
 
Molarity of CaCl2 ,
 
No. of moles of CaCl2 
 
equals 11 over 111

equals 0.099 space mol
 
Molarity 
 
equals fraction numerator No. space of space moles over denominator Vol space in space straight L end fraction

equals fraction numerator 0.099 over denominator 0.5 end fraction

space equals 0.1981 space straight M
 
Molar conductance of CaCl2 
 
 
straight capital lambda subscript straight m space equals fraction numerator straight K cross times 1000 over denominator Molarity end fraction

space space space space space space equals fraction numerator 7.83 cross times 10 to the power of negative 4 end exponent cross times 1000 over denominator 0.1981 end fraction

space space space space space space equals 3.96 space ohm to the power of negative 1 end exponent space mol to the power of negative 1 end exponent cm squared

Answered by Varsha | 14th Aug, 2019, 11:46: AM