What is the vapor pressure at 20 degree C of an ideal solution prepared by the a

Question

What is the vapor pressure at 20 degree C of an ideal solution prepared by the addition of 8.87 g of the nonvolatile solute urea, CO(NH_2)_2, to 57.6 g of methanol. CH_3OH? The vapor pressure of pure methanol at 20 degree C is 89.0 mmHg. Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. K_t = 1.86 degree C/m Dimethylglyoxime, DMG, is an organic compound used to test for aqueous nickel(II) ions. A solution prepared by dissolving 65.0 g of DMG in 375 g of ethanol boils at 80.3 degree C. What is the molar mass of DMG? K_b = 1.22 degree C/m, boiling point of pure ethanol = 78.5 degree C Calculate the molecular weight of a small protein if a 0.24-g sample dissolved in 108 ml. of water has an osmotic pressure of 9.5 mmHg at 22 degree C. (R = 0.0821 L middot atm/(K middot mol)) What concentration of aqueous FeCl_3 would have the same osmotic pressure as a 0.20 M solution of CaCl_2 at the same temperature, assuming ideal behavior?

Explanation / Answer

(a)

According to raoult's law,

P0 -P / P0 = n1/ n1 +n2

(89 - P) /89 = (8.87/60) / ((8.87/60)+(57.6/32))

P = 82.24 mm Hg

(b)

Use the formula for freezing point depression.
Change in T = -Kf x molality of solution
In this solution, molality (b) = 0.31 mol sucrose / .175 kg water = 1.8 moles sucrose / kg water.
Therefore,
Change in T = - 1.86 C/m (1.8 molals sucrose) = -3.3 C
Normally, water freezes at 0 C. But with sucrose added to the solution, the freezing point drops to about -3.3 C

(c)

We know that dT b = Kb * m
Where
?T f = elevation in boiling point
= boiling point of solution - boiling point of pure ehtanol
= 80.3 - 78.5
= 1.8 oC
K b = elevation in boiling point constant = 1.22 oC / m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= ( 65g / M ) / 0.375 Kg
Plug the values we get 1.8 = 1.22 * ( 65 / M ) / 0.375 )
M = 117.48 g / mol

(d)

9.5 /760 atm = M * 0.08206 L atm mol-1K-1 * (273+22)

M = 9.5 /760 atm / 0.08206 L atm mol-1K-1 * (273+22)

M = 5.16* 10^-4 mol L-1

moles of protein = 5.16* 10^-4 mol L-1* 108mL = 0.5573* 10^-4 moles

molecular weight of protein = mass of protein / moles

           = 0.24/0.5573* 10^-4

= 4.3 * 10^3 g/mol

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