Asked by Ken
Al2(s04)3 solution of 1 molal concentration is present in 1 litre solution of 2.684g/cc.How many moles of bas04 would be precipitation on adding bacl2 in excess?
The answer 6M
I do not understand pls explain step by step
Thank a lot all tutors
The answer 6M
I do not understand pls explain step by step
Thank a lot all tutors
Answers
Answered by
bobpursley
Step by step.
The first sentence. Weird. It says you have one M solution. Lets figure that.
1M=masspresent/(volume*molmass)
1M*molmass:*1Liter=mass
1*342.15gram=mass
or mass/cc=.342 g/cc
So why does it then state 2.684g/cc ? Nuts to this problem, your teacher is not thinking.
Finally, you aske how many moles of BaSO4 there would be, and you tell me the answer is 6 Molal? Nuts again.
The first sentence. Weird. It says you have one M solution. Lets figure that.
1M=masspresent/(volume*molmass)
1M*molmass:*1Liter=mass
1*342.15gram=mass
or mass/cc=.342 g/cc
So why does it then state 2.684g/cc ? Nuts to this problem, your teacher is not thinking.
Finally, you aske how many moles of BaSO4 there would be, and you tell me the answer is 6 Molal? Nuts again.
Answered by
DrBob222
I agree the problem makes no sense. I have brought this up before to Ken; m stands for molal while M sands for molar.
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