A 40.0% C_6H_12O_6(aq) has a density of 1.15 g/mL at 18 degree C a. What is the

Question

A 40.0% C_6H_12O_6(aq) has a density of 1.15 g/mL at 18 degree C a. What is the molarity of the solution? b. What is the molality of the solution? c. What is the mole fraction of solvent in the solution d. Estimate the normal boiling point of the solution(K_b of H_2O = 0.512 degree C/m; normal boiling point of H_2O = 100.00 degree C e. Estimate the freezing point of the solution (K_f of H_2O = 1.86 degree C/m; freezing point of H_2O = 0.00 degree C f. Estimate the osmotic pressure of this solution at 18 degree C g. Estimate the vapor pressure of this solution at 18 degree C. The vapor pressure of pure H_2O at this temperature is 17.5 torr

Explanation / Answer

Basis:1 g of solution

0.4 g of C6H12O6, 0.6 g of water.

Volume of solution=mass of solution/density of solution=1 g/1.15 g /mL=0.8695 mL

Moles of solute C6H12O6=Mass of solute(0.4 g)/molar mass( 180.16 g/mol)=0.0022 mol

Molarity=moles of solute*1000/volume in mL=0.0022 mol*1000/0.8695 L =2.553 M

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kg of solvent =0.6g =0.6*10^-3kg

Molality=moles of solute/kg of solvent=0.0022 mol*1000/0.6=3.633 mol/kg=3.633m

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Moles of solvent=0.6g/18 g/mol =0.033 mol

Total moles=0.033+0.0022=0.0352 mol

mole fraction of solvent water=0.033 mol /0.0352 mol=0.9375

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Change in boiling point delta T=Kb*molality=0.512 C/m*3.633=1.86 C

Delta T=normal boiling point of solution-boiling point of water

normal boiling point of solution=1.86 C+100 C=101.86 C

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