Question

How much benzoate tartrate (M. wt. = 774.8) is required to make 1200 ml  of a 8 % solution of benzoate (M.wt = 285.3) please show steps

Answer 1

8% solution of benzoate is " 8 g. of benzoate present in 100 mL of solution.

i.e. In 1200 mL of solution, the weight of benzoate present = (1200/100)*8 = 96 g

The no. of moles of benzoate = 96 g/(285.3 g/mol) = 0.336488 mol

Therefore, the weight of benzoate tartrate required = 0.336488 mol * 774.8 g/mol = 260.71 g