Saved Bookmarks
| 1. |
Calculte the degree of hardness of a sample of water containing `6 mg` of `MgSO_(4)` per kg of water. |
|
Answer» Correct Answer - 5 `because 1 g-"mole"` or `120 g MgSO_(4) "has" =1 g-"mole"` or `100g CaCO_(3)` `therefore 6xx10^(-3)g MgSO_(4) "has" =(100xx6xx10^(-3))/(120)` `=5xx10^(-3)gCaCO_(3)` Thus `1000 g` of water contains `MgSO_(4)` equivalent to `5xx10^(-3)gCaCO_(3)` `therefore 10^(6)g` water contains `=(5xx10^(-3))/(1000)xx10^(6)` `5 g "of" CaCO_(3)` |
|