CBSE Class 12-science Answered
Plz answer 3.14
![question image](https://images.topperlearning.com/topper/new-ate/topr_81134271272386483rps20190813132635.jpeg)
Asked by lovemaan5500 | 13 Aug, 2019, 17:47: PM
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](https://images.topperlearning.com/topper/tinymce/cache/583af4a76527e196c613fdb5ad73a536.png)
=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](https://images.topperlearning.com/topper/tinymce/cache/8439eb60951424c438d845457be81b9b.png)
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](https://images.topperlearning.com/topper/tinymce/cache/34b9b004195c2260f3e061ffab5f0304.png)
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](https://images.topperlearning.com/topper/tinymce/cache/4017cba9593148374ca9a45b1cbd7637.png)
![fraction numerator No. space of space gram space equivalents over denominator Volume space of space solution space in space straight L end fraction](https://images.topperlearning.com/topper/tinymce/cache/4017cba9593148374ca9a45b1cbd7637.png)
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](https://images.topperlearning.com/topper/tinymce/cache/bdb7a23e427f91bb0bf0c51f4a0e604c.png)
Normality of CaCl2 =
![equals fraction numerator 0.1981 over denominator 0.5 end fraction
equals 0.3963 space straight N](https://images.topperlearning.com/topper/tinymce/cache/f8093b442cadd031f1613052807ff3be.png)
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](https://images.topperlearning.com/topper/tinymce/cache/dd748f13bf5fa65d5e37d30871144f47.png)
Molar conductance of CaCl2 =
Molarity of CaCl2 ,
No. of moles of CaCl2
![equals 11 over 111
equals 0.099 space mol](https://images.topperlearning.com/topper/tinymce/cache/0262c87fe6c7e27f8eabdcbb862621c9.png)
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](https://images.topperlearning.com/topper/tinymce/cache/bf4b8bf8a820b90c61dcc9a0a09a4158.png)
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](https://images.topperlearning.com/topper/tinymce/cache/87c44e96a97da4ce75cfa5c90283ae6c.png)
Answered by Varsha | 14 Aug, 2019, 11:46: AM
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