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प्रश्न
Calculate \[\ce{\wedge^0_m BaCl2 and Al2(SO4)}\] from the following data.
For \[\ce{\wedge^0_m Ba^2+ = 127.2 S cm^2 mol^{-1}, \wedge^0_m Al^3+ = 189 S cm^2 mol^{−1}}\]\[\ce{\wedge^0_m Cl= 76.3 S cm^2 mol^{-1}, \wedge^0_m SO^2-_4 = 160 S cm^2 mol^{-1}}\]
उत्तर
\[\ce{\wedge^0_m (BaCl2) = \wedge^0_m (Ba^{+2})+ 2\wedge^0_m (Cl^-)}\]
= 127.2 + 2 × 76.3
= 279.8 S cm2 mol−1
\[\ce{\wedge^0_m(Al2(SO4)3) = 2\wedge^0_m(Al^{+3}) + 3\wedge^0_m(SO^2-_4)}\]
= 2 × 189 + 3 × 160
= 858 S cm2 mol−1
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