A7.06% aqueous solution of sodium bicarbonate has a density of 1.19 g/ml at 25°c

what is the molarity of the solution

0.706 m

0.0840 m

0.904 m

1.00 m

a 7.06% aqueous solution of sodium bicarbonate has a density of 1.19 g/ml at 25°c

what is the molality of the solution

0.904 m

1.00 m

0.0840 m

0.706 m

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

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

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

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

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 ∘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 °c

79.1°c

97.6 °c

78.5°c

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.

-22.5 °c

-23.34 °c

-17.71 °c

-28.13 °c

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'd be really indebted to anyone who could me with this

Q1:

M = 1.0 mol/L.

Explanation:

•To solve this problem, we can use the relation:

M = 10Pd / Molar mass, where

M is the molarity of the solution (mol/L),

P is the percent of the solution (P = 7.06 %),

d is the density of the solution (d = 1.19 g/ml),

Molar mass of sodium bicarbonate (84.007 g/mol),

M = 10 (7.06 %) (1.19 g/ml) / (84.007 g/mol) = 1.0 mol/L.

Q2:

Molality = 0.904 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

The solute is sodium bicarbonate (7.06 %), it is the component with the lower percent.

The solvent is water (92.94 %), it is the component with the higher percent.

•To get the mass of the solute and solvent, we should firstly get the mass of the solution.

Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (1.19 g/ml) (1000 ml) = 1190 g.

The mass of solute sodium bicarbonate = (the total mass of the solution x percent % of solute) / 100 = (1190 x 7.06) / 100 = 84.014 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (1190 x 92.94) / 100 = 1105.986 g.

m = (mass / molar mass) solute x (1000 / mass of solvent)

m = (84.014 g / 84.007 g/mol) (1000 / 1105.986 g) = 0.904 m.

Q3:  

Mole fraction of solvent (water) = 0.953.

Explanation:

•The solute is methanol CH3OH (8.05 %), it is the component with the lower percent.

•The solvent is water (91.95 %), it is the component with the higher percent.

•To solve this problem, we can use the relation of mole fraction of water (X water):

X water = (no. of moles of water) / (total no. of moles of the solution)

•To get the no. of moles of the solute and solvent, we should obtain the mass of both.

•To obtain the mass of the solute and solvent, we should firstly get the mass of the solution.

•Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (0.976 g/ml) (1000 ml) = 976.0 g.

The mass of solute methanol CH3OH = (the total mass of the solution x percent % of solute) / 100 = (976.0 x 8.05) / 100 = 78.568 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (976.0 x 91.95) / 100 = 897.432 g.

The no. of moles of solute methanol CH3OH = mass / molar mass = (78.568 g) / (32.0 g/mol) = 2.45525 mol.

The no. of moles of solvent water = mass / molar mass = (897.432 g) / (18.0 g/mol) = 49.857 mol.

X water = (no. of moles of water) / (total no. of moles of the solution) = (49.857) / (49.857 + 2.45525) = 0.953.

Q4:

m = 5.18 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

Mass of the solvent water = 60.0 g.

m = (mass / molar mass) cesium chloride x (1000 / mass of water)

m = (52.3 g / 168.36 g/mol) (1000 / 60.0 g) = 5.177 m = 5.18 m.

Q5:

4.91 M.

Explanation:

•To solve this problem, we can use the relation of molarity (M):

M = (mass / molar mass) solute x (1000 / V of the solution)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

V of the solution = 63.3 ml.

M = (mass / molar mass) cesium chloride x (1000 / V of the solution)

M = (52.3 g / 168.36 g/mol) (1000 / 63.3 g) = 4.907 M = 4.91 M.

Q6:

The vapor pressure of the solution = 80.0 torr.

Explanation:

•The vapor pressure of the solution will be the partial vapor pressure of the ethanol only because alpha naphthol is assumed to be nonvolatile, so it has no partial vapor pressure in the total vapor pressure of the solution.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol),  

Where, X ethanol is the mole fraction of ethanol.

P0 ethanol is the vapor pressure of pure ethanol (P0 = 100.0 torr).

•To get the mole fraction of ethanol, we can use the relation of mole fraction of ethanol (X ethanol):

X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution).

•The no. of moles of ethanol = mass / molar mass = (36.8 g) / (46.07 g/mol) = 0.80 mol.

•The no. of moles of alpha naphthol = mass / molar mass = (28.8 g) / (144.0 g/mol) = 0.2 mol.

•X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution) = (0.80) / (0.80 + 0.2) = 0.80.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol) = (0.80) (100.0 torr) = 80.0 torr.

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