Need help with these 3 questions please show all work thanks

n: (medium) What is the freezing point of an aqueous aluminum nitrate solution freezing point depressio that is 10.0 % aluminum nitrate by mass? Assume complete dissociation. Answer: °C (medium) 10.0 g of a non-electrolyte was dissolved in 115.0 g of benzene (CaHs) resulting in a solution that freezes at 0.70 °C. If the normal freezing point of benzene is 5.53 °C and its Kr-5.12 °C/m, calculate the molar mass of the solute. Answer:g/mol osmotic pressure: (easy) Assuming complete dissociation, what mass of iron(III) chloride (molar mass 162.20 gmo) needs to be added 255 mL of water at 35°C to cause an osmotic pressure of 260 mmHg? Answer: mg

Solution

5.

Suppose mass of solution = 100 g

Mass of water = 100-10 = 90 g = 0.090 Kg

Mass of aluminium nitrate = 10 g

Number of moles of aluminium nitrate = mass/molar mass = 10/213 = 0.04695

Molality (m) of solution = number of moles of aluminium nitrate/mass of solvent in Kg

= 0.04695/0.090 = 0.5216

Since, aluminium nitrate is completely dissociate, so, vant Hoff factor i = 4

Al(NO₃)₃ = Al³⁺ (aq) + 3NO₃^-

deltaT = i x K_f x m = 4 x 1.86 x 0.5216 = 3.881 deg C

Freezing point of solution = 0 -3.881 = - 3.881 deg C

6.

Mass of benzene = 115 g = 0.115 Kg

Mass of solute = 10 g

Number of moles of solute = mass/molar mass = 10/MW

Molality (m) of solution = number of moles of solute/mass of solvent in Kg

= 10/(0.115 x MW)

deltaT = i x K_f x m

(5.53 – 0.70) = 1 x 5.12 x 10/(0.115 x MW)

4.83 = 51.2/(0.115 x MW)

MW = 51.2/(0.115 x 4.83) = 92.18 g/mole

7.

Suppose mass of FeCl3 added = W g

Number of moles of FeCl3= mass/molar mass = W/162.20

Volume of water = 225 mL = 0.225 L

Molarity of solution (M) = number of moles of FeCl3/Volume of solution in L

= (W/162.20)/0.225 = W/(0.225 x 162.20) = W/36.495

Since, FeCl3 is completely dissociate, so, vant Hoff factor i = 4

FeCl₃ = Fe³⁺ (aq) + 3Cl^-

Osmotic pressure = 2.60 mmHg = 2.60/760 atm = 0.00342 atm

P = i M RT

0.00342 = 4 x (W/36.495) x 0.082 x (35 + 273.15)

36.495 x 0.00342 = 4 x W x 0.082 x 308.15

W = 1.23 mg