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0.1 M-KMnO_(4) is used for the following titration. How much volume of the solution in ml will be required to react with 0.158 gm of Na_(2)S_(2)O_(3)? S_(2)O_(3)^(2-)+MnO_(4)^(-)+H_(2)OtoMnO_(2)(s)+SO_(4)^(2-)+OH^(-) |
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Answer» 80 ML `(0.1xxV_(mL))/(1000)xx3=(0.158)/(158)xx8impliesV_(mL)=26.67mL` |
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