Grade 12Physical Chemistry
density of a 2.05M solution of acetic acid in water is 1.02g/mL.The molality of the solution is???
4 Answers
Happy Man
2
Vikas TU
Let us find the weight of the acetic acid
= 2.05 * 60
=123
The weight of the solution
=1000* 1.02
=1020
Weight of the water =(1020-123)
=897g
Therefore, Molarity is given as;
=2.05 *1000897
=2.285molkg-1
Ritik
We can solve it easily by this formula-d=M[1/m+M`/1000] Where d=density of solution, M=molarity of substance, m= molality of substance, M`=Molecular weight of substance.
Rishi Sharma
Dear Student,
Please find below the solution to your problem.
Suppose the volume of solution is 1000 ml
1ml=1.2gm
1000ml=1020gm
Number ofr moles =2⋅05M×1lit
=2.05mol
Mass of solute =n×M⋅wt
=2.05×60
=123gm
Solvent mass =1020−123
=897gm
Molality =123×1000/60×897
=2.28molkg−1
Thanks and Regards
Please find below the solution to your problem.
Suppose the volume of solution is 1000 ml
1ml=1.2gm
1000ml=1020gm
Number ofr moles =2⋅05M×1lit
=2.05mol
Mass of solute =n×M⋅wt
=2.05×60
=123gm
Solvent mass =1020−123
=897gm
Molality =123×1000/60×897
=2.28molkg−1
Thanks and Regards
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