CBSE Class 12-science Answered

Plz answer 3.14

Asked by lovemaan5500 | 13 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 | 14 Aug, 2019, 11:46: AM

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