A 8.05 % CH3OH(aq) has a density of 0.976 g/mL at 18°C What is the mole fraction of the solvent in the solution? 0.953 8.05 91.95 0.0469

2) An aqueous solution of cesium chloride is prepared by dissolving 52.3 g cesium chloride in 60.0g of water at 25°C. The volume of this solution is 63.3 mL . What is the molality of the solution?
0.311 m
4.91 m
2.77 m
5.18 m
3) An aqueous solution of cesium chloride is prepared by dissolving 52.3 g cesium chloride in 60.0g of water at 25°C. The volume of this solution is 63.3 mL . What is the molarity of the solution?
4.91 M
2.69 M
5.18 M
2.77 M

4) The vapor pressure of ethanol, C2H5OH is 100.0 torr at 35 °C. Calculate the vapor pressure of the solution formed by dissolving 28.8 g of alpha naphthol, C10H8O, in

36.8 g of C2H5OH. Assume alpha naphthol to be nonvolatile at this temperature.

20.0 torr
43.9 torr
80.0 torr
56.1 torr

5) Both ethanol, C2H5OH and propanol, C3H7OH, are volatile. At 35 °C, the vapor pressure of pure ethanol is 100 torr and that of propanol is 37.6 torr. What is the vapor pressure at this temperature of a solution is formed by mixing 36.9 g of ethanol and 12.0 g propanol.
15.3 torr
50.1 torr
84.7 torr
87.5 torr

The boiling point of pure ethanol, C2H5OH, is 78.4 LaTeX: ^\circ ∘ C. Its boiling point elevation constant is 1.22 °C/m. What is the boiling point of a solution formed by dissolving 8.00 g of alpha-naphthol (C10H7OH) in 100.0 g ethanol.

91.3 degrees centigrade
79.1 degrees centigrade
97.6 degrees centigrade
78.5 degrees centigrade


The freezing point of CCl4 is -22.92°C. Calculate the freezing point of the solution prepared by dissolving 17.5g of pyrazine (C4H4N2) in 1250g of CCl4. The freezing point depression constant for CCl4 is 29.8 °C/m.
-22.50
-23.34
-17.71
-28.13


Consider the following aqueous solutions:

A. 0.10 M NH4NO3,

B. 0.10 M Fe(NO3)3

C. 0.10 M Ba(NO3)2

D. 0.10 M NH2CONH2

Arrange the following in increasing order (smallest to largest) order of osmotic pressure

C < B < A < D
A < D < C < B
D < A < C < B
A < C < B < D

Consider the following aqueous solutions:

A. 0.10 m NH4NO3,

B. 0.10 m Fe(NO3)3

C. 0.10 m Ba(NO3)2

D. 0.10 m NH2CONH2

Arrange the following in increasing order (smallest to largest) order of Freezing point. The freezing point of pure water is 0.00 ∘ C and its freezing point depression constant is 1.86 ∘ C/m

D < A < C < B
B < C < A < D
C < B < A < D
A < C < B < D

Consider the following aqueous solutions:

A. 0.10 m NH4NO3,

B. 0.10 m Fe(NO3)3

C. 0.10 m Ba(NO3)2

D. 0.10 m NH2CONH2

Arrange the following in increasing order (smallest to largest) order of normal boiling point. The normal boiling point of pure water is 100.00 ∘ C and its boiling point elevation constant is 0.512 ∘ C/m

C < B < A < D
D < A < C < B
A < C < B < D
B < C < A < D

A solution is prepared by dissolving 1.22 g of compound in enough water to make up 262 mL in volume. The osmotic pressure of the solution is found to be 30.3 torr at

35.0 °C. Calculate the molar mass of the compound.

257 g/mol
2950 g/mol
3.88 g/mol
44.7 g/mol
i tried to solve them all but i keep getting wrong answers can anyone help

Respuesta :

Mass of solute, W_B = 8.05 g[/tex]

Molar mass of solute, M_B = 32 g /mol

Mass of solvent,[tex]W_A = 91.95 g[/tex]

Molar mass of solvent,[tex]M_A = 18 g /mol[/tex]

Number of mol of solvent is

[tex]n_A = W_A /M_A[/tex]

[tex]= 91.95 / 18 = 5.108 mol[/tex]

Number of mol of solute is

n_B = W_B /M_B

= 8.05 / 32 = 0.25 mol

Mole fraction of solvent is the number of moe of solvent divided by total nmber of mole of solute and solvent as follows:

[tex]X_A = n_A /(n_A + n_B)[/tex]

[tex]= 5.108 / (5.108 + 0.25) = 0.954[/tex]

Thus, mole fraction of solvent is 0.954