Saved Bookmarks
| 1. |
Calculate the molarity of a solution of CaCl_(2) if on chemical analysis it is found that 200 ml of CaCl_(2) solution contains 3.01xx10^(22) chloride ions. |
|
Answer» Solution :`underset("1 mole")(CaCl_(2))rarrCa^(2+)+underset(2xx6.02xx10^(23)" IONS")(2Cl^(-))` `therefore 3.01xx10^(22)Cl^(-1)"ions will be present in CaCl"_(2)=(1)/(2xx6.02xx10^(23))xx3.01xx10^(22)" mole" = "0.025 mole"` `therefore"MOLARITY of solution "=("0.025 mole")/("0.200 L")=0.125M.` |
|