Solution,
molarity is the no of moles of solute per unit volume.
We can calculate amount of CaCl2 required to prepare 0.1 M CaCl2 1000 ml solution.
we know that to prepare one ltr of 1 M solution of CaCl2 111 g required
Now consider x gram will require to prepare to
so that comparing above both condition
1000ml ×1M×X g=1000ml×0.1M×111g
X= 11.1 gram
X= 11.1 g of CaCl2
Hence 11.1 g of CaCl2 would be dissolved in 1.0L of a 0.100 M solution of CaCl2
