1. what is the vapor pressure at 20 degrees celsius of a solution of 3.50g of naphthalene C10H8 in 25.0g of benzene (v.P pure benzene = 74.7 mm Hg at 20 degrees celsius)

2. calculate the freezing point of a solution of 25 mL of CH3OH (d= 0.792 g/mL) in 325 mL of water (d=1.00g/mL, Kf= 1.86 degrees celcius/m).

3. a 5.00g sample of laurylalcohol is discovered in 100g of benzene (Kf= 5.10 degrees celcius/m). the normal feezing point of benzene is 5.50 degrees celcius while the solution froze at -1.78 degrees celcius. what is the molecular wieght of lauryl alcohol?

4. a compound contains 42.9% C, 2.4% H, 16.6% N, and 38.1% O by mass. the addition od 3.16g of this compound to 75 mL of cyclohexane (d= 0.779g/ml) gives a solution with a freezing point of 0.0 degrees celsius. the normal freezing pt of cyclohexane is 6y.50 degrees celcius and its freezing pt depression constant of 20.2 degrees celcius/m. what is the molecular formula of the solute.




Attempts:
moles C10H8 = 3.50 g/128.17 g/mol=0.0273
moles benzene = 25.0 g/ 78.11 g/mol=0.320

mole fraction benzene = 0.320 / 0.320 + 0.0273 = 0.921
mole fraction C10H8 = 0.0273 / 0.320 + 0.0273=0.0786
p benzene = 0.921 x 74.4 = 68.5 mm Hg
I have found v .P naphthalene at 25 °C = 0.078 mm Hg ; using this value
p C10H8 = 0.078 x 0.0786 =0.00613 mm Hg
p = 0.00613 + 68.5 = 68.5 mm Hg

mass CH3OH = 25 x 0.792 = 19.8 g
moles = 19.8 / 32.042 g/mol=0.618
m = 0.618 / 0.325 Kg = 1.90
delta T = 1.90 x 1.86=3.54
f.p. = - 3.54 °C

delta T = 5.50 - ( - 1.78)=7.28
7.28 = m x 5.10
m = 1.43 = moles solute / 0.100 Kg
moles solute = 0.143
molar mass = 5.00/ 0.143 =35.0 ( this is not true : check your data)

are these correct attempts to begin?

Q1 and Q2 are done correctly

Q3 method is correct but I am not sure why you are getting wrong molar mass
May be data used is not correct.