JEE MAIN - Chemistry (2005 - No. 36)
| Electrolyte: | KCl | KNO3 | HCl | NaOAc | NaCl |
|---|---|---|---|---|---|
| $${ \wedge ^\infty }(Sc{m^2}mo{l^{ - 1}}):$$ |
149.9 | 145 | 426.2 | 91 | 126.5 |
517.2
552.7
390.7
217.5
Explanation
$$A_{HCl}^\infty = 426.2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$A_{AcONa}^\infty = 91.0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$$A_{NaCl}^\infty = 126.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
$$A_{AcOH}^\infty = \left( i \right) + \left( {ii} \right) - \left( {iii} \right)$$
$$ = \left[ {426.2 + 91.0 - 126.5} \right]$$
$$ = 390.7$$
$$A_{AcONa}^\infty = 91.0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$$A_{NaCl}^\infty = 126.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
$$A_{AcOH}^\infty = \left( i \right) + \left( {ii} \right) - \left( {iii} \right)$$
$$ = \left[ {426.2 + 91.0 - 126.5} \right]$$
$$ = 390.7$$
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