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