10 grams of sucrose was used to prepare a 0.5 molar solution. a) Approximately how much water (in mL was added)? b) What is the molality of this solution? c) What is the mole fraction of the water in the solution?

2 Answers
Jan 1, 2018

Sucrose's molecular formula is #C_12H_22O_11#, thus its atomic weight is approximately #(342g)/("mol")#. Hence,

#10g * ("mol")/(342g) * 1/(xL) = 0.5M#
#therefore x approx 58.5mL#

To check our work,

#10g * ("mol")/(342g) * 1/(0.0585L) approx 0.5M#

For the next part, we must assume the density of water is approximately #(1g)/(mL)#. Hence,

#(10g * ("mol")/(342g))/(58.5mL * (1g)/(mL) * (kg)/(10^3g)) approx 0.5m#

The reason these may be the same is because in the first part I assumed water was the only part of the solution that added volume. Likewise, it is the solvent. Therefore, with molality, paying only attention to the solvent, we have the same mass. Perhaps I went wrong here, and I'm open to feedback if so.

Moving on with the third part, mole fraction is just what is sounds like,

#(58.5g * (H_2O)/(18g))/(10g * (C_6H_12O_6)/(342g) + 58.5g * (H_2O)/(18g)) approx 0.99#

There are no units in this case, since all of them cancel out.

Jan 1, 2018

a) The volume of water added was 52 mL; b) #b = "0.56 mol/kg"#; c) #chi_text(water) = 0.990#

Explanation:

Step 1. Calculate the moles of sucrose

#n_text(sucrose) = 10 color(red)(cancel(color(black)("g"))) × "1 mol"/(342.30 color(red)(cancel(color(black)("g")))) = "0.0292 mol" #

Step 2. Calculate the volume of the solution

#V = 0.0292 color(red)(cancel(color(black)("mol"))) × "1 L"/(0.5 color(red)(cancel(color(black)("mol")))) = "0.058 L = 58 mL"#

Step 3. Calculate the mass of the solution

The density of a 0.5 mol/L solution of sucrose is 1.064 g/mL.

#m = 58 color(red)(cancel(color(black)("mL"))) ×"1.064 g"/(1color(red)(cancel(color(black)("mL")))) = "62 g"#

Step 4. Calculate the volume of the water

#"Mass of water = (62 - 10) g = 52 g"#

#"Volume of water" = 52 color(red)(cancel(color(black)("g"))) × "1.0 mL"/(1 color(red)(cancel(color(black)("g")))) = "52 mL"#

Step 5. Calculate the molal concentration of the sucrose

#b = "moles of solute"/"kilograms of solvent" = "0.0292 mol"/"0.052 kg" = "0.56 mol/kg"#

Step 6. Calculate the mole fraction of water

#n_text(water) = 52 color(red)(cancel(color(black)("g"))) × "1 mol"/(18.02 color(red)(cancel(color(black)("g")))) = "2.9 mol"#

#chi_text(sucrose) = n_text(sucrose)/(n_text(sucrose) + n_text(water)) = (0.0292 color(red)(cancel(color(black)("mol"))))/((0.0292 + 2.9) color(red)(cancel(color(black)("mol")))) = 0.0292/2.9 = 0.010#

#chi_text(water) = 1 - chi_text(sucrose) = 1 - 0.010 = 0.990#