The molality of a 1 litre solution with x percent H^2SO4 is 9.The weight of solvent present in the solution is 910 grams. The value of x:
Answers
We have approx. 50% H2SO4 ww........
Explanation:We are given a concentration in terms of molality.
molality=Moles of soluteKilograms of solute=9⋅mol⋅kg−1.
And thus....................................
Moles of solute=9⋅mol⋅kg−1×0.910⋅kg=8.19⋅mol, the which represents 8.19⋅mol×98.08⋅g⋅mol−1=803.28⋅g with respect to sulfuric acid.
And thus...........
%sulfuric acid m/m=mass of sulfuric acidmass of solution=803.28⋅g803.28⋅g+910⋅g×100%
=46.9%=D as required.......
Do you see what I have done here? We were given the molality, and then we used this value to calculate the percentage by mass. The total mass included the sulfuric acid and the water...Quite clearly, we also had to KNOW definitions of molality, mol⋅kg−1, and also percentage by mass......mass of solutemass of solute + mass of solvent×100%.