Determine the molarity, molality, mole fraction, mass fraction, boiling point, f

Question

Determine the molarity, molality, mole fraction, mass fraction, boiling point, freezing point, and vapor pressure at 25 C (where the vapor pressure of pure water is 23.756 torr) of a 0.16% (w/w) solution of potassium sulfate.

1. A tree is 10 m tall. If the sap rises due to osmotic pressure at 20 C, what must be the total molarity of the solutes in the sap? Assume the groundwater is pure water with a density of 1.0 g/mL and that 1 mmHg 13.6 mmH2O. 2. The simplest way to figure out the van't Hof factor, i, is the count the ions that a species dissociates into. The dissociation, however, is affected by the concentration of solutions with more concentrated solutions having lower amounts of dissociation (a) Calculate the van't Hoff factor for a 2.00 (w/w) H2SO4 solution that freezes at -0.796 °C. (b) How does this help us decide whether a sulfuric acid solution is made up of mostly H2SO4, of mostly HSO4 and H30, or of mostly 2 H30+ and SO2 3. Determine the molarity, molality, mole fraction, mass fraction, boiling point, freezing point, and vapor pressure at 25 oC where the vapor pressure of pure water is 23.756 torr) of a 0.16% (w/w) solution of potassium sulfate

Explanation / Answer

Molarity:

weight of K2SO4 dissolved = 0.16 g
volume of solution = 100g = 100 mL
gram mol weight = 174 g (GMW)

Molarity M = (G/GMW)x (1000/ v in mL) = (0.16/174)* (1000/100) = 0.0092 M

molality :

0.16 % solution means 0.16 g of K2SO4 present in 100g of water;
so, weight of water = 100-0.16 = 99.84 g
molality = (G/GMW)x (1000/ a in g) = (0.16/174)* (1000/99.84) = 0.0092 m

mole fraction:

weight of K2SO4 = 0.16 g; mol wt = 174
no. of moles K2SO4 = n(s) = 0.16/174 = 0.0009

weight of water = 99.84 g; mol.wt = 18
no. of moles H2O n(o) = 99.84/18 = 5.5466

Mole fraction of K2SO4 X(s) = n(s)/(n(s) + n(o)) = 0.0009/ (0.0009 + 5.5466) = 0.00016

Mass fraction;

mass of K2SO4 = m(s) = 0.16 gram
mass of H2O = m(o) = 99.84 gram

Mass fraction of K2SO4 = Xm(s)= m(s)/ m(total) = 0.16/100 = 0.0016

Boiling point :
t = m • Kb • i

m = molality of solution; (Note : i =3; for K2SO4)

Kb = not given?

Freezing point :

t = m•Kf • i

Kf = not given;

Vapour pressure ;

P = Xsol x Po(solvent)

P = (5.5466/(0.0009 + 5.5466)) x 23.756 torr = 23.752 torr

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.