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Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm^{3} and 34% solute by mass. Calculate the molarity. (1mL = 1 cm^{3})

1.19 g/cm^{3} ---> 1190 g / L

( 1190 g / L ) x ( 34g /100g ) = 404 g/ L

(404 g/ L) / (36.46 g HCL/ mol) = 11.09 mol/ L

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO_{3}), density 1.42 g/cm^{3} and 65% solute by mass. Calculate the molarity.

1.42 g/cm^{3} ---> 1420 g / L

( 1420 g / L ) x ( 64g /100g ) = 908.8 g/ L

(908.8 g/ L) / (63 g HNO_{3}/ mol) = 14.7 mol/ L

Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm^{3} and 31% solute by mass. Calculate the molality.

moles HCl = 31 g / (36.461 g/mol) = 0.850 mol

mass water = 100 - 31=" 69" g = 0.069 Kg

m = .850 / 0.069 =12.32 mol / kg

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO_{3}), density 1.42 g/cm^{3} and 67% solute by mass. Calculate the molality.

moles HNO_{3}= 67 g / ( 63 g/mol) =1.06 mol

mass water = 100 - 67=" 33" g = 0.033 Kg

m = 1.06 / 0.033 = 32.21 mol / kg

Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm^{3} and 31% solute by mass. Calculate the mole fraction.

moles HCl = 31g / (36.461 g/mol) = 0.850 mol

mass water = 100 - 31 = 69 g

moles water = 69 g/ 18.02 g/mol= 3.82 mol

mole fraction HCl = .850 / (0.850 + 3.82) = 0.18

Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO_{3}), density 1.42 g/cm^{3} and 61% solute by mass. Calculate the mole fraction.

moles HCl = 61 g / (63 g/mol) = 0.968 mol

mass water = 100 - 61 = 39 g

moles water = 39 g/ 18.02 g/mol= 3.385

mole fraction HCl = .968 / (0.850 + 3.82) = 0.31

On the basis of the nature of the intermolecular attractions, arrange water, ethanol (C_{2}H_{5}OH), and chloroform (CHCl_{3}) in order of decreasing solubility in carbon tetrachloride (CCl_{4}).

Select one:

- Nonpolar with nonpolar (like dissolves likes)
- CHCl = nonpolar
- C
_{2}H_{5}OH = polar (only one polar bond O-H) - H
_{2}O = polar (two polar bonds O-H )

Predict the relative solubility of each compound in the two solvents on the basis of intermolecular attractions. Is NaCl more soluble in water or in carbon tetrachloride?

Select one:

- NaCl is ionic compound (polar) that completely dissolves in water

Is iodine (I_{2}) more soluble in water or in benzene?

Select one:

- Iodine (I
_{2}) is nonpolar so it dissolves in a nonpolar solution C_{6}H_{6. (like dissolves like)}

Is ethanol (C_{2}H_{5}OH) more soluble in water or in hexane?

Select one:

- Ethanol (C
_{2}H_{5}OH) is polar so it dissolves in polar solution (

Is ethylene glycoL (HOCH_{2}CH_{2}OH) more soluble in ethanol or in benzene?

Select one:

- Ethylene glycoL (HOCH
_{2}CH_{2}OH) is a polar so it dissolves in a polar solution ( ). The O-H bond makes any molecule polar.

Check the correct statements regarding solubilities in liquids. (The following figure for question (c) shows the solubility of some ionic compounds as a function of temperature.)

Select one or more:

**Rule of Thumb:** As **temperatures increase** **gases decrease in solubility**

The steeper slope the greater solubility with increase in temperature.

At 25 ^{o}C and 2.0 atm, the enthalpy of solution of neon in water is -2.46 kJ/mol, and its solubility is 9.07 x 10^{-4} m. Check the conditions for which the solubility of neon is greater than 9.07 x 10^{-4}m.

Select one or more:

Low temp and high pressure increases the solubility of Ne.

The enthalpy of solution of nitrous oxide (N_{2}O) in water is -12 kJ/mol and its solubility at 20 ^{o}C and 1.00 atm is 0.121 g per 100. g of water. Calculate the molal solubility of nitrous oxide in water at 0.800 atm and 20 ^{o}C. Hint, first find Henry's law constant at 20 ^{o}C and 1.00 atm.

Hint given in feedback

Hint given in feedback

Solubility = 0.121 g per 100 g of water

Assuming density of water ≈ 1 g/mL, volume of solution = 100 mL

Concentration, c (molarity) = mass * 1000 / (molar mass * volume of solution)

= 0.121 g * 1000 / (44.01 g/mol * 100 mL)

= 0.0275 M

Pressure, P = 1.00 atm

Henry's law constant, k = c/ P

k = 0.0275 M/ 1.00 atm

k = 0.0275 M/atm

Solubility, c = k * P

= 0.0275 M/atm * 0.800 atm

= 0.022 M

A 0.511 g sample of a nonvolatile, yellow crystalline solid dissolves in 13.0 g of benzene, producing a solution that freezes at 5.39 ^{o}C. Find the molar mass of the yellow solid. The following may be useful: The freezing point of benzene is 5.51 ^{o}C and the freezing point depression constant, k_{f}, is 4.90 ^{o}C/m.

Hint given in feedback

Hint given in feedback

Procedure:

(1) Solve ΔT_{f} = mk_{f} for m and find m.

(2) Recall m = mol solute/ kg solvent.

(3) Multiply m by **kg** solvent to get moles solute (yellow solid).

(4) Recall the definition of molar mass is g/mol. Simply divide the mass of the solute by the number of moles this mass corresponds to.

Δt = i K_{f} m

Pure benzene freezes at 5.51 °C

Δt = 5.51 -5.39 = 0.12

0.12 °C = (1) (4.9°C kg mol^{-1}) (x / 0.013 kg)

x = 3.183 x 10^{-4} Mol

**Molar mass of the compound = 0.511 g/ 3.183 x 10 ^{-4} mol = 1605 gm/mol**

A 4.60% solution by mass of an enzyme in water has an osmotic pressure of 17.9 torr at 286 K. Calculate the molar mass of this enzyme. (Assume that the density of the solution is 1.00 g/mL.)

Hint given in feedback

Hint given in feedback

Procedure:

(1) Solve π = MRT for M and find M.

(2) Recall M = mol solute/L solution.

(3) Multiply M by L solution to get moles solute.

(4) Recall the definition of molar mass is g/mol. Simply divide the mass of the solute by the number of moles this mass corresponds to.

M = π / RT

R = 62.36367 L torr/K mol

M = 17.9 / ( 62.36*286) = 0.0010036 mol/L

.0460g x 1g/ml x 1000ml/L = 46 g/L

( 46 g/L) / ( 0.0010036 mol/L ) = 45832.86 g/mol = 46000 g/mol

A saline solution contains 6.1 g of NaCl in 1.00 kg of water. Assuming an ideal value for the Van't Hoff factor, i, calculate the freezing point of this solution in ^{o}C.

6.1 g of NaCl

6.1g/ (58 g/mol NaCl) = 0.10436 mol NaCl

m = 0.10436 mol / 1kg = 0.10436

K = 1.86 (water)

i = 2 (NaCl dislocates into two ions)

T_{ f} = (0.10436) (1.86) (2) = 0.388

0 - 0.388 = -0.388

A saline solution contains 7.1 g of NaCl per 0.80 liter of solution. Assuming an ideal value for the van't Hoff factor, i, calculate the osmotic pressure (in atm) of this solution at 291 K.

i = 2 (NaCl dislocates into two ions)

R = .08206T = 291 K

M = molality (mol / L)

7.1 g NaCl / (58.45 g/mol NaCl) = 0.1215 mol NaCl

0.1215 mol NaCl / 0.80L = 0.15187

π = 2 x 0.15187 x .08206 x 291 K = 7.25 torr

A reverse osmosis unit is used to obtain drinkable water from a source that contains 0.600 g NaCl per liter. What is the minimum pressure (in torr) that must be applied across the semipermeable membrane to obtain water? Assume room temperature or about 298 K. (Don′t forget the van′t Hoff factor!)

i = 2 (NaCl dislocates into two ions)

R = 62.364 L Torr molT = 298 K

M = (.6 g/L) x (1 mol/ 58.45g NaCl) = .010265 mol/L

π = 2 x .010265 mol/L x 62.364 x 298 K = 381 torr

When 0.030 mol of HCl dissolves in 100.0 g of benzene, the solution freezes at 4.04 ^{o}C. When 0.030 mol of HCl dissolves in 100.0 g of water, the solution freezes at -1.07 ^{o}C. Use data in the following table to calculate the expected molality of HCl in each solvent, ignoring the van't Hoff factor. Check the 3 statements which are correct.

Select one or more:

delta T in benzene is 5.50 - 4.0 = 1.5 C

molality is (0.03 mol / 0.1 kg) = 0.3 m

1.5 = (5.12)(0.3)(1)

The van't Hoff factor is equal to 1 which means that HCl does NOT dissociate in benzene.

delta T in water is 1.07 C

molality is still 0.3 m

1.07 = (1.86)(0.3)(2)

The van't Hoff factor is equal to 2 which means that HCl DOES dissociate in water.

A mixture contains 25 g of cyclohexane (C_{6}H_{12}) and 44 g of 2-methylpentane (C_{6}H_{14}). The mixture of liquids is at 35 ^{o}C . At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase?

C_{6}H_{12 = 25g / (84 g/mol) = 0.2976 mol}

(C_{6}H_{14}) = 44g / (86 g/mol) = 0.5116 mol

0.2976 / (0.2976 + 0.5116) = 0.3677

What is the vapor pressure (in torr) of cyclohexane above the solution?

P_{ cyc vapor}= x P_{ cyc liquid}

P_{ cyc vapor}= 0.3677 (150)

P_{ cyc vapor}= 55.5

What is the mole fraction of cyclohexane in the vapor phase?

Calculate Pi for C6H14 first

55.5 / (55.5 + 197.9) = 0.22

At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 ^{o}C) and water (H_{2}O; boiling point = 100.0 ^{o}C) form an azeotropic mixture, boiling at 107.1 ^{o}C, that is 77.5% by mass formic acid. At the boiling point of the azeotrope (107.1 ^{o}C), the vapor pressure of pure formic acid is 917 torr, and that of pure water 974 torr. If the solution obeyed Raoult's law for both components, what would be the vapor pressure (in torr) of formic acid at 107.1 ^{o}C?

77.5 g / (45g/mol HCOOH) = 1.68

22.5 g/ (18.02g/ mol H2O) = 1.25

mole fraction of acid: 1.68 / (1.25+1.68) = 0.573

P_{ acid vapor } = x P_{ acid liquid}

_{P acid vapor = (0.573) x (917 torr) }

_{P acid vapor = 525.79 torr}

If the solution obeyed Raoult's law for both components, what would be the vapor pressure of the water (in torr) at 107.1 ^{o}C?

mole fraction of water: 1- 0.573 = 0.427

P_{ acid vapor } = x P_{ acid liquid}

_{P acid vapor = (0.427) x (974torr) }

_{P acid vapor = 415.53 torr}

If the solution obeyed Raoult's law for both components, what would be the total vapor pressure (in torr) at 107.1 oC?

P_{ total } = P _{acid } + P_{ water }

_{}

P_{ total }= 941.32 (theoretical)

Check those statements which are true regarding the above azeotropic mixture.

Select one or more:

The calculated pressure is the theoretical value. The actual value is less.

A negative deviation means actual value is less than the theoretical. This means the solutes and solvents have a strong bond.