Question

Question 1 :

a) Show the calculation of the mass of Al2(SO4)3 needed to make 200ml of a 0.250 M solition. report your answer to 3 significant figures.

b) show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NO3)2

c)show the calcultion of the freezing point of a solution made by dissolving 15.6 grams of the nonelectrolyte C3H6O3 in 200 geams of water. Kf for water is 1.86, freezing point of pure water is 0C. calculate hour answer to 0.01C

d) show the calculation of the molar mass (molecular weight) of a solute if a solution of 13.9 grams of solute in 200 grams of water has a freezing point of -1.40C. Kf for water is 1.86 and the feeezing point of pure water is 0C. calculafe your answerr to 0.1g/mole.

Answer 1

a)

molar mass of Al₂(SO₄)₃ = 342.15 g/mol

Volume of the solution = 200 mL

molarity of the solution = 0.250 M

1000 massof solute molarity volumeofsolution(mL) molarmassofthesolute

massof solute 1000 0.250M 200m L 342.15g/mol

massofsolute 0.250M x 342.15g/mol 200mL 1000

massof solute 17.1075g

b)

molarity of the solution = 0.667 M

mass of the solute = 37.5 g

molar mass of solute = 261.3 g/mol

1000 massof solute molarity volumeofsolution(mL) molarmassofthesolute

37.5g 1000 0.667M 261.3g/mol volumeofsolution (m L)

37.5g 1000 wolumeofsolution (mL) 0.667M 261.3g/mol

volumeo fsolution (mL) 215.16mL

c)

mass of solute ( C₃H₆O₃ ) = 15.6 g

mass of solvent (water) = 200 g

K_f = 1.86

molar mass of solute ( C₃H₆O₃ ) = 90 g/mol

depression in freezing point = K_f $\times$ molality of solution

massof solute 1000 molality X massof solvent(g) molarmassof thesolute

1000 15.6g molality90q/mol 200g X

depression in freezing point = K_f $\times$ molality of solution

depression in freezing point = 1.86$\times$ 0.867

depression in freezing point = 1.61262⁰ C

depression in freezing point = 1.61⁰ C (rounding off to two decimals)

freezing point = normal freezing point - depression in freezing point

freezing point = 0⁰ C - 1.61⁰ C

freezing point = - 1.61⁰ C

d)

mass of solute = 13.9 g

mass of solvent (water) = 200 g

freezing point = -1.40⁰C

normal freezing point = 0⁰C

freezing point = normal freezing point - depression in freezing point

depression in freezing point= 0⁰C - ( -1.40⁰C )

depression in freezing point= 1.40⁰C

K_f = 1.86

depression in freezing point = K_f $\times$ molality of solution

1.40⁰C = 1.86$\times$  molality of solution

1.40 C molalityof solution 1.86

molality of solution = 0.753 m

massof solute 1000 molality X massof solvent(g) molarmassof thesolute

$0.753 m =\frac{13.9 g}{molar mass of the solute}\times \frac{1000}{200 g}$

$molar mass of the solute =\frac{13.9 g}{0.753 m}\times \frac{1000}{200 g}$

molarmassofthesolute 92.29g/mol