Answer:
The answer is 2, 12 g of LiCl in 1 L of solution
Explanation:
We calculate the weight of 1 mol of LiCl:
Weight 1 mol LiCl= Weight Li + Weight Cl =6, 94 g + 35, 45 g= 42,39 g/mol
In a 0.05M solution it means that we have 0.05 moles of compound (in this case, LiCl) in 1 liter of solution:
1 mol---------42, 39 g LiCl
0,05mol-----x=(0,05molx 42, 39 g LiCl)/ 1 mol =2, 1195 g LiCl
