please can someone work this problem and show step by step a

please can someone work this problem and show step by step and also please write clearly so I can understand thank you.

Page 10 (VIII) (10 points) At 298 K, the osmotic pressure of a glucose (CeHi20) solution is 12.94 atm. Calculate the freezing point of the solution. The density of the solution is 1.16 g/ml.(For water, the normal freezing point is 0.00°C and Kr value is 1.86°C/m.)

Solution

Given

Kf = 1.86 C/m , T = 298 K , osmatic pressure( )= 12.94 atm and Density of slution = 1.16 g/mL

R = 0.0821 lit atm K^-1 mol^-1 (from literature) molar mass of glucose =180.15 g/mol

According to von`t hoff equation

V= nRT rearrange this as = nRT / V   ( we know that n/ V = Molarity M)

so equation become as

= MRT

12.94 = M x 0.0821 x 298

M = 0.5289 mol / L

So mass of glucose = num of moles x molar mass of glucose =0.5289 x 180.15 =95.28 gm

So mass of glucose = 95.28 gm (solute )

Now calculate molality of solution

Assume that we have given 1 L of solution this contain 0.5289 mole of glucose

So mass of 1 L solution = 1000 mL x density = 1000 x 1.16 = 1160 gm

mass of 1 L solution = 1160 gm

mass of water = mass of solution - mass of solute = 1160 - 95.28 =1064.72 gm (mass of water )

mass of water=1064.72 gm or 1.06472 kg

Molality = num of mole of solute per kg of solvent

Molality = 0.5289 / 1.06472 =0.497 m

Now according to freezing point depression

Tf = molar depression constant (Kf) x molality

Tf = 1.86 x 0.497 =0.92C

Tf =0.92C

so the given solution of glucose freezes at 0.92C