Indicate how many particles are formed when the following solutes dissolve. SOLU
ID: 695110 • Letter: I
Question
Indicate how many particles are formed when the following solutes dissolve. SOLUTE sucrose (Ci2HzOn) sodium sulfate (Na SO.) # OF PARTICLES SOLUTE magnesium chloride (MgCI) methanol (CH OH) # OF PARTICLES Calculate the freezing point of a solution of 50.0 g methyl salicylate, CyH602, dissolved in 800. g of benzene, CoHl6- Ke for benzene is 5.10°C/m and the freezing point is 5.50°C for benzene. 2.89 C The molal boiling point constant for ethyl alcohol is 1.22 /molal. Its boiling point is 78.4 C. A solution of 14.2 g of a nonvolatile nonelectrolyte in 264 g of the alcohol boils at 798C. What is the molecular mass of the solute A solution of 0.2113 g of water dissolved in 25.0 g of a solvent freezes at 11.5°C below the freezing point of the solvent. What is Kf for this solvent? 24.5 C/m 125 g of the non-volatile solute glucose, CalhO is dissolved in 125 g of water at 25.0°C. IF the vapor pressure of water at 25.0°C is 23.7 Torr, what is the vapor pressure of the solution? Dang3Explanation / Answer
We know that T f = Kf x m
Where
T f = depression in freezing point
= freezing point of pure solvent – freezing point of solution
= 5.50-Tf oC
K f = depression in freezing constant= 5.10C/m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
=(50/138)/0.8kg
= 0.453 m
Plug the values we get Tf= 3.20oC
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We know that T b = Kbx m
Where
T b= elevation in boiling point
= boiling point of solution - boilinging point of pure solvent
= 79.8-78.4=1.40 oC
Kb = elevation in boiling point constant of water = 1.22 oC/m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= (14.2/ M) / 0.264kg
= 53.8/M
Plug the values we get 46.9 g/mol
Do the remaining in similar manner