Asked by lucy
If 25cm^3 of NaOH solution containing 4.0g per litre of solution neutralised 50cm^3 of a monobasic acid,HX containing 1.8g of acid.Calculate the R.F.M of the acid. I got the molarity of NaOH to be 0.1,then i got stuck.plz help
Answers
Answered by
bobpursley
RFM...relative formula mass
Ma*Va=Mb*Vb
Vb=.025 Va=.050
Ma=Mb* .5
Mb= 4/40=.1 so
Ma=.05
but Molarity acid= mass/RFM*Va= 1.8/RFM*.060=36/RFM
RFM=36/.05=720 check my math.
Ma*Va=Mb*Vb
Vb=.025 Va=.050
Ma=Mb* .5
Mb= 4/40=.1 so
Ma=.05
but Molarity acid= mass/RFM*Va= 1.8/RFM*.060=36/RFM
RFM=36/.05=720 check my math.
Answered by
DrBob222
Interesting. I showed Lucy how to work this problem a day or so ago. I guess she wanted someone to do the work for her.
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