What is the molarity and molality of a 13 solution
Asked by admin @ in Chemistry viewed by 296 People
calculate the molarity and molality of 13% solution by weight of sulphuric acid (density is equal to 1.020 gm)
Asked by admin @ in Chemistry viewed by 296 People
calculate the molarity and molality of 13% solution by weight of sulphuric acid (density is equal to 1.020 gm)
Answered by admin @
Answer:
M=~10.20M, m=1.524m
Explanation:
Here,
weight of sulphuric acid=13g(13% out of 100g solution)
Molecular weight=Mw=98
volume=?
density= mass/volume
volume=mass/density
v=13/1.020
v=12.74ml~13
Now,
Molarity(M)= weight of solute×1000
Mw × V(ml)
=13×1000/98×13
=1000/98
=~10.20M
Now,
weight of solute=13g.
We know that,
weight of solution=weight of solute+weight of solvent.
100=13+weight of solvent.
100-13=weight of solvent
87g=weight of solvent.
Now, we know that,
Molality(m) = weight of solute ×1000
Mw × weight of solvent(in g)
= 13×1000/98×87
=13000/98×87
=13000/8526
=1.524m
Thank you.
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