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What is the electrical conductivity of a solution containing 0.1 M sodium chloride and 0.2 M potassium nitrate at room temperature (25 °C) and a cell constant of 1.0 cm-1?

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To calculate the electrical conductivity of the solution, we need to consider the contributions of each ion in the solution. The electrical conductivity    can be calculated using the following equation: =  ci * i where ci is the concentration of each ion  in mol/L  and i is the molar conductivity of each ion  in S cm/mol .First, let's identify the ions and their concentrations in the solution:1. Sodium ion  Na+ : 0.1 M  from sodium chloride 2. Chloride ion  Cl- : 0.1 M  from sodium chloride 3. Potassium ion  K+ : 0.2 M  from potassium nitrate 4. Nitrate ion  NO3- : 0.2 M  from potassium nitrate Next, we need the molar conductivity values    for each ion at 25 C. These values can be found in tables or literature. Here are the approximate values for each ion:1.  Na+  = 50.1 S cm/mol2.  Cl-  = 76.3 S cm/mol3.  K+  = 73.5 S cm/mol4.  NO3-  = 71.4 S cm/molNow, we can calculate the electrical conductivity    using the equation mentioned above: =  0.1 M * 50.1 S cm/mol  +  0.1 M * 76.3 S cm/mol  +  0.2 M * 73.5 S cm/mol  +  0.2 M * 71.4 S cm/mol  =  5.01 S cm  +  7.63 S cm  +  14.7 S cm  +  14.28 S cm  = 41.62 S cmSince the cell constant is given as 1.0 cm-1, we can directly use the calculated  value as the electrical conductivity of the solution:Electrical conductivity of the solution = 41.62 S/m

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