Question

can anyone help me with this?

The density of an aqueous solution containing 12.50 g K2SO4 in 100.00 g solution is 1.083 g/mL. Calculate the concentration of this solution in terms of molarity, molality, percent of K2SO4, and mole fraction of solvent.

i don't know how to get the liters of solution

Answer 1

solute-K2SO4
My solution on Molarity is
M=moles of solute/liters of solution

moles of K2SO4= 12.50g/174.2592g/mol
= 0.0717 mol.

M= 0.0717mol/ Liters of solution

Answer 2

100 g solution ( 1 mL/1.083 g) = ? mL solution?

Answer 3

Density = mass/volume
so
Volume = mass/density

Answer 4

You've got 100.00 g of solution with density 1.083 g/mL, so the volume of solution is:
(100.00 g) / (1.083 g/mL) = 92.336 mL

(12.50 g K2SO4) / ( 174.2592 g K2SO4/mol) / (0.092336 L) = 0.7769 mol/L K2SO4

(12.50 g K2SO4) / ( 174.2592 g K2SO4/mol) / (100.00 g) x (1000 g) = 0.7173 m

(12.50 g K2SO4) / (100.00 g) = 12.50% K2SO4 by mass

(100.00 g total) - (12.50 g K2SO4) = 87.50 g H2O
(87.50 g H2O) / (18.01532 g H2O/mol) = 4.85698 mol H2O
(12.50 g K2SO4) / ( 174.2592 g K2SO4/mol) = 0.0717322 mol K2SO4
(4.85698 mol) / (4.85698 mol + 0.0717322 mol) = 0.9854 mol fraction of H2O

Answer 5

thank you :)