Question

calculate the amount of benzoic acid required to prepare 250
mL of 0.15M solution in methanol ​

Answer 1

Answer:

number of moles of benzoic acid required =0.15×250/1000=0.0375moles

molar mass of benzoic acid=7 (12)+6(1)+2 (16)=84+6+32=122g/mil

So mass of benzoic acid required =122×0.0375=4.575g