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Calculate wedge_(m)^(0)" for "CaCl_(2) and MgSO_(4). lambda_(Ca^(2+))^(0)=119.0S cm^(2)" mol"^(-1) lambda_(Cl^(-))^(0)=76.3S cm^(2) mol^(-1) lambda_(Mg^(2+))^(0)=106 Scm^(2) mol^(-1) lambda_(SO_(4)^(2-))^(0)=160 Scm^(2) mol^(-1) |
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Answer» Solution :`wedge_(m)^(0) (MgSO_(4))=lambda_(Mg)^(0)+lambda_(SO_(4)^(2-))^(0)` `=106+160=266 S cm^(2) MOL^(-1)` `wedge_(m)^(0) (CaCl_(2))=lambda_(Ca^(2+))^(0)+2lambda_(Cl^(-))^(0)=119+2 xx 76.3 =271.6Scm^(2) mol^(-1)` |
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