What is the molar concentration of cesium cation in 245 mL aqueous solution containing 0.15 g cesium carbonate?
3 years ago
So this problem can be approached through the following steps:
1. Calculate the number of moles of Cs2CO3 from the given mass. This is the moles present in 245 mL of aqueous solution
2. Calculate the number of moles of Cs+ cation present in 245 mL of the solution. 1 mol Cs2CO3 contains 2 moles of Cs+ cation.
3. Calculate the moles of Cs+ in 1000 mL of the solution which will be the molar concentration of Cs+ cation in M (mol/L or molarity).
Step 1: Number of moles of Cs2CO3 = 0.15/Molecular weight of Cs2CO3 = 0.15/326 = 0.00046 mol
Step 2: 1 mol Cs2CO3 contains 2 mol Cs+ cation. Hence, 0.00046 mol Cs2CO3 contains 0.00092 mol Cs+
Step 3: 0.00092 Cs+ cation is present in 245 mL of the solution. Hence, in 1000 mL of solution, 0.00037 mol of Cs+ will be present.
Thus, molar concentration of Cs+ cation is 0.00037 mol/L of solution.
3 years ago
Answered By Kaushik S
So this problem can be approached through the following steps:
1. Calculate the number of moles of Cs2CO3 from the given mass. This is the moles present in 245 mL of aqueous solution
2. Calculate the number of moles of Cs+ cation present in 245 mL of the solution. 1 mol Cs2CO3 contains 2 moles of Cs+ cation.
3. Calculate the moles of Cs+ in 1000 mL of the solution which will be the molar concentration of Cs+ cation in M (mol/L or molarity).
Step 1: Number of moles of Cs2CO3 = 0.15/Molecular weight of Cs2CO3 = 0.15/326 = 0.00046 mol
Step 2: 1 mol Cs2CO3 contains 2 mol Cs+ cation. Hence, 0.00046 mol Cs2CO3 contains 0.00092 mol Cs+
Step 3: 0.00092 Cs+ cation is present in 245 mL of the solution. Hence, in 1000 mL of solution, 0.00037 mol of Cs+ will be present.
Thus, molar concentration of Cs+ cation is 0.00037 mol/L of solution.