Calculate the normal boiling point of a solution consisting of 1.0 gram of naphthalene dissolved in 100.0 grams of toluene. The normal boiling point of toluene is 110.70C and the boiling point constant is 36.1 K
Lets calculate molality first
Molar mass of C10H8,
MM = 10*MM(C) + 8*MM(H)
= 10*12.01 + 8*1.008
= 128.164 g/mol
mass(C10H8)= 1.0 g
use:
number of mol of C10H8,
n = mass of C10H8/molar mass of C10H8
=(1 g)/(1.282*10^2 g/mol)
= 7.803*10^-3 mol
m(solvent)= 100 g
= 0.1 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(7.803*10^-3 mol)/(0.1 Kg)
= 7.803*10^-2 molal
lets now calculate ΔTb
ΔTb = Kb*m
= 36.1*0.078
= 2.8167 oC
This is increase in boiling point
boiling point of pure liquid = 110.7 oC
So, new boiling point = 110.7 + 2.8167
= 113.5167 oC
Answer: 113.52 oC
Calculate the normal boiling point of a solution consisting of 1.0 gram of naphthalene dissolved in 100.0 grams of tolue...
9. Calculate the normal boiling point of a solution consisting of 1.0 gram of naphthalene dissolved in 100.0 grams of toluene. The normal boiling point of toluene is 110.7°C and the boiling point constant is 36.1 K. (6 pts)
Calculate the freezing point and boiling point of a solution containing 10.0 g of naphthalene (C10Hz) in 100.0 mL of benzene. Benzene has a density of 0.877 g/cm².
How many grams of sulfur (S8) must be dissolved in 100.0 naphthalene (C10H8) to make a 0.12 m solution?
Calculate the boiling point of a solution of 10.0 grams of picric acid, C6H2(OH)(NO2)3(s), dissolved in 100.0 grams of cyclohexane, C6H12(s). Assume that the colligative molality and the molality are the same for picric acid in cyclohexane. not sure if more information is needed. this is all i got!
What is the boiling point of a 2.31 m solution of naphthalene in benzene? The normal boiling point of benzene is 80.1C and the Kb for benzene is 2.53 C/m)
The boiling point of water is 100.0°C at 1 atmosphere. How many grams of manganese(II) bromide (214.7 g/mol), must be dissolved in 270.0 grams of water to raise the boiling point by 0.300°C? Refer to the table for the necessary boiling or freezing point constant Solvent Formula Water H20 Ethanol CH3CH2OH Chloroform CHCI: Benzene CH Diethyl ether CH2CH OCH CH3 K (°C/m) Kr(°C/m) 0.512 1.86 1.22 1.99 3.67 2.53 5.12 2.02 g manganese(II) bromide.
Calculate the freezing point and boiling point of a solution containing 17.4 g of naphthalene (C10H8) in 112.0 mL of benzene. Benzene has a density of 0.877 g/cm3. Part A Calculate the freezing point of a solution. (Kf(benzene)=5.12∘C/m.) Part B Calculate the boiling point of a solution. (Kb(benzene)=2.53∘C/m.)
Calculate the freezing point and boiling point of a solution containing 20.0 g of naphthalene (C_10H_8) in 119.0 mL of benzene. Benzene has a density of 0877 g/cm^3. Calculate the freezing point of a solution (K_f (benzene) = 5.12 degree C/m) T_f = 2.54 Calculate the boiling point of a solution. (K_b (benzene) = 2.53 degree C/m) T_b =
What are the boiling point and freezing point of a 4.97 m solution of naphthalene in benzene? (The boiling point and freezing point of benzene are 80.1 C and 55 degree C respectively. The boiling point elevation constant for benzene is 2.53 C/m, and the freezing point depression constant for benzene is 5.12 degree C/m.'')
The boiling point of water is 100.0°C at 1 atmosphere. How many grams of aluminum nitrate (213.0 g/mol), must be dissolved in 279.0 grams of water to raise the boiling point by 0.450°C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kb (°C/m) Kf (°C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02