Question

A. Determination of Solubility Above Ambient Temperature Temperature 24°C Mass of apparatus 117. 33 g Mass of apparatus with oxalic acid dehydrate 123.33 Mass of oxalic acid dihydrate used 6 g Mole fraction Crystal formation Temperature (°C) Concentration mass solute/100 H2O Total Volume (mL) 10 ml S m mu 480 45°C 417°C 10 ml 24° C B. Determination of Solubility below Ambient Temperature mass of apparatus. Temperature 117.338 Mass of oxalic acid dihydrate used 1. 89 Total Volume Crystal formation Concentration Mole fraction (mL) Temperature (°C) mass solute/100 g H20 10 ml 30°C 15 ml 34°C 20 ml 38°C Show your calculations for the solution concentrations and mole fractions. ncneil

please heeelp me compute the concentration mass solute/100 g H2O and mole fraction please. i don't know how to do it. thank you

Answer 1

Student need help in computing mass solute/100 g H₂O and mole fraction :

A.

Mass of Solute (Oxalic Acid Dihydrate ) = 6.00 g

Molar mass of Oxalic Acid Dihydrate = 126 g /mol

Density of Oxalic Acid Dihydrate = 1.65 g/cm³

So, volume of Oxalic Acid Dihydrate used = 6.00 g/ 1.65 g/cm³ = 3.64 cm3 = 3.64 ml

(1 cm3 = 1 ml)

1. Total Volume = 10 ml ; so, volume of water : 10-3.64 = 6.36 ml

density of water = 1g/ ml

so, mass of water = 6.36 g

Concentration solute (mass solute/100 g H₂O ) = ( 6.00 g / 6.36 g ) *100 g

= 94.33 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate = 6.00g /126 g /mol = 0.0476 mole

Moles of water =6.36g /18 g /mol = 0.353 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0476 mole / 0.4006 mole = 0.119

Similarly ,

2. For  Total Volume = 15 ml   

concentration solute (mass solute/100 g H₂O ) = ( 6.00 g / 11.36 g ) *100 g

= 52.82 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate = 6.00g /126 g /mol = 0.0476 mole

Moles of water =11.36g /18 g /mol = 0.631 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0476 mole / 0.6787 mole = 0.07

3. For  Total Volume = 20 ml   

concentration solute (mass solute/100 g H₂O ) = ( 6.00 g / 16.36 g ) *100 g

= 36.67 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate = 6.00g /126 g /mol = 0.0476 mole

Moles of water = 16.36g /18 g /mol = 0.909 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0476 mole / 0.9566 mole = 0.05

B.

Mass of Solute (Oxalic Acid Dihydrate ) = 1.8 g

Molar mass of Oxalic Acid Dihydrate = 126 g /mol

Density of Oxalic Acid Dihydrate = 1.65 g/cm³

So, volume of Oxalic Acid Dihydrate used = 1.8 g/ 1.65 g/cm³ = 1.09 cm3 = 1.09 ml

(1 cm3 = 1 ml)

1. Total Volume = 10 ml ; so, volume of water : 10 -1.09 = 8.91 ml

density of water = 1g/ ml

so, mass of water = 8.91 g

Concentration solute (mass solute/100 g H₂O ) = ( 1.8 g / 8.91 g ) *100 g

= 20.2 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate =1.8 g /126 g /mol = 0.0143 mole

Moles of water =8.91g /18 g /mol = 0.495 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0143 mole / 0.5093 mole = 0.028

Similarly ,

2. For  Total Volume = 15 ml   

Concentration solute (mass solute/100 g H₂O ) = ( 1.8 g / 13.91 g ) *100 g

= 12.94 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate =1.8 g /126 g /mol = 0.0143 mole

Moles of water =13.91g /18 g /mol = 0.773 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0143 mole / 0.787 mole = 0.018

3. For  Total Volume = 20 ml   

Concentration solute (mass solute/100 g H₂O ) = ( 1.8 g / 18.91 g ) *100 g

= 9.52 g / 100 g H₂O

Moles of Oxalic Acid Dihydrate =1.8 g /126 g /mol = 0.0143 mole

Moles of water = 18.91g /18 g /mol = 1.05 mole

mole fraction (x) of solute = moles of solute / total moles in solution

= 0.0143 mole / 1.0648 mole = 0.0134