Molarity Molality and Mole Fraction Key - Chemistry 302

[Pages:7]Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page ? find all choices before answering.

ChemPrin3e G 13 001 (part 1 of 2) 10.0 points Determine the mass of anhydrous copper(II) sulfate that must be used to prepare 250 mL of 1.9 M CuSO4(aq).

Correct answer: 75.8148 g.

Explanation: V = 250 mL = 0.25 L

M = 1.9 M

003 10.0 points A chemist studying the properties of photographic emulsions needed to prepare 500 mL of 0.178 M AgNO3(aq). What mass of silver nitrate must be placed into a 500 mL volumetric flask, dissolved, and diluted to the mark with water?

Correct answer: 15.1187 g.

Explanation: v = 500 mL = 0.5 L

M = 0.178 M

FWAgNO3 = 107.8682 g/mol + 14.0067 g/mol + 3 (15.9994 g/mol)

= 169.873 g/mol

FWCuSO4 = 63.546 g/mol + 32.066 g/mol + 4 (15.9994 g/mol)

= 159.61 g/mol

mAgNO3 = (0.178 M) (0.5 L) ? (169.873 g/mol)

= 15.1187 g

mCuSO4 = (1.9 M) (0.25 L) (159.61 g/mol) = 75.8148 g

Brodbelt 013 015

004 10.0 points How much NaNO3 is needed to prepare 225 mL of a 1.55 M solution of NaNO3?

002 (part 2 of 2) 10.0 points Determine the mass of CuSO4 ? 5 H2O that must be used to prepare 250 mL of 1.9 M CuSO4(aq).

Correct answer: 118.603 g. Explanation:

FWCuSO4?5 H2O = FWCuSO4 + 5 (FWH2O) = 159.61 g/mol + 10 (1.0079 g/mol) + 5 (15.9994 g/mol) = 249.69 g/mol

mCuSO4?5 H2O = (1.9 M) (0.25 L) (249.69 g/mol) = 118.603 g

1. 4.10 g

2. 29.6 g correct

3. 12.3 g

4. 0.132 g

5. 0.244 g Explanation: V = 225 mL

M = 1.55 M

?

g

NaNO3

=

225

mL

?

1 L soln 1000 mL

1.55 mol NaNO3 ? 1 L soln

85 g NaNO3 ? 1 mol NaNO3

= 29.6 g NaNO3

ChemPrin3e G 05

Brodbelt 03 13

Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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005 10.0 points How many moles of HCl are present in 40.0 mL of a 0.035 M solution?

1. 0.012 mol

2. 0.0060 mol

3. 0.25 mol

4. 0.0012 mol

5. 0.0014 mol correct

Explanation: V = 40.0 mL

M = 0.035 M

? mol HCl = 40.0 mL soln

1L

0.035 mol HCl

? 1000 mL ? 1 L soln

= 0.0014 mol HCl

Brodbelt 013 515 006 10.0 points What is the final concentration of Ca(OH)2 when 255 mL of 0.250 M Ca(OH)2 is mixed with 55.0 mL of 0.65 M Ca(OH)2?

1. 0.390 M

in the two individual solutions:

? mol Ca(OH)2 (soln 1)

= 0.255 L soln

0.250 mol Ca(OH)2

?

1 L soln

= 0.06375 mol Ca(OH)2

? mol Ca(OH)2 (soln 2)

= 0.055 L soln

0.65 mol Ca(OH)2

?

1 L soln

= 0.03575 mol Ca(OH)2

Total mol Ca(OH)2 = 0.06375 mol + 0.03575 mol

= 0.0995 mol Ca(OH)2

The total volume of the final solution is the sum of the volumes of the individual solutions.

Total L = 0.255 L + 0.055 L = 0.31 L soln Molarity is moles solute per L of solution.

?

M

Ca(OH)2

=

0.0995 mol Ca(OH)2 0.31 L soln

= 0.321 M Ca(OH)2

Msci 14 0813

007 10.0 points What is the effective molality of a solution containing 12.0 g of KI in 550 g water? Assume 100 percent ionic dissociation.

2. 0.642 M

1. 0.072 molal

3. 0.780 M

2. 0.26 molal correct

4. 2.90 M

3. 0.42 molal

5. 0.900 M

4. 0.066 molal

6. 0.321 M correct

5. 0.13 molal

Explanation:

V1 Ca(OH)2 = 255 mL [Ca(OH)2]1 = 0.250 M

V2 Ca(OH)2 = 55 mL [Ca(OH)2]2 = 0.65 M The total moles of Ca(OH)2 in the final

solution will be the sum of the moles present

6. 0.59 molal

Explanation:

mKI = 12.0 g

mH2O = 550 g

Because KI is a 1:1 salt, you get one cation

and one anion for every single formula unit

Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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that dissolves. Therefore, you'll get a DOUBLING of the stated molality for the effective molality. The formula weight of KI is 166 g/mol, so the number of moles of KI is

12.0 g KI

1 mol KI 166 g KI

= 0.0723 mol KI ,

and the (stated) molality of the solution would then be

0.0723 mol KI 0.550 kg H2O

=

0.131

m.

But recall that the effective molality would be twice the stated molality here, so the effective molality is 0.262 m.

Nlib 11 0020 008 10.0 points If you mix 3 moles of ethylene glycol (antifreeze) in 4165 grams of water, what is the molality of the solution?

Correct answer: 0.720288 m.

Explanation:

nethylene glycol = 3 mol

mwater = 4165 g

Molality (m) is moles solute per kilogram

of solvent. The solute is ethylene glycol. The

solvent is water and 4165 g = 4.165 kg.

?

m

=

3

mol ethylene glycol 4.165 kg H2O

=

0.720288

m

Nsci 14 0808 009 10.0 points 1.9 g of NaCl and 6.1 g of KBr were dissolved in 48 g of water. What is the molality of NaCl in the solution?

Holt da 13 1 sample 1 010 10.0 points

What is the molarity of a 3.047 L solution that is made from 11.29 g of NaCl?

Correct answer: 0.0634032 M.

Explanation:

Vsolution = 3.047 L M =?

mNaCl = 11.29 g

M

=

mol L

solute soln

? mol NaCl = 11.29 g NaCl

1 mol NaCl ? 58.44 g NaCl

= 0.19319 mol NaCl

?

M

=

mol NaCl L soln

=

0.19319 mol 3.047 L

= 0.0634032 M

Molality 08 44a 011 10.0 points Calculate the molality of sucrose in a solution composed of 11.31 g of sucrose (C12H22O11) dissolved in 606 mL of water.

Correct answer: 0.054524 m.

Explanation: mC12H22O11 = 11.31 g VH2O = 606 mL = 0.606 L

MWC12H22O11 = 342.296 g/mol

Correct answer: 0.6773 m.

Explanation: mNaCl = 1.9 g mwater = 48 g

mKBr = 6.1 g

mH2O = (0.606 L)

1 kg 1L

Thus the molality is

= 0.606 kg

mNaCl

=

mol NaCl kg water

1.9 g

=

58.4 g ? mol-1 NaCl 0.048 kg water

= 0.6773 m

mC12 H22 O11

=

moles solute kg solvent

11.31 g sucrose

=

342.296 g/mol sucrose 0.606 kg H2O

= 0.054524 m

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Msci 14 0809 012 10.0 points What additional information, if any, would enable you to calculate the molality of a 7.35 molar solution of a nonelectrolyte solid dissolved in water?

1. Both the density of the solution and the molecular weight of the solute would be needed. correct

2. Only the density of the solution would be needed.

d = 0.9344 g/cm3

Assume 1 L of 10.5 M NH3(aq); it will contain 10.5 mol NH3 with a mass of

(10.5 mol)(17.0305 g/mol) = 178.82 g .

The density of the 1 L of solution is

0.9344

g/cm3

?

1000 cm3 1L

=

934.4

g/L ,

so the total mass of the solution is 934.4 g, which leaves 934.4 g - 178.82 g = 755.58 g of water.

Therefore, the molality is

3. Only the molecular weight of the solute would be needed.

m

=

10.5 mol 0.75558 kg

NH3 solvent

=

13.8966

m

.

4. Only the density of water would be needed.

5. None is needed. Explanation:

molarity

=

mol solute L solution

molality

=

mol solute kg solvent

The density of the solution can be used to convert volume (1 L) of solution into mass of solution. Then the molecular weight of the solute (given or calculated from the formula) can be used to convert the number of moles solute in 1 L solution into mass of solute in grams. The mass of the solvent is the difference between the mass of the solution and the mass of the solute (both of which have been calculated). Substitute the values into the molality formula and calculate.

Molality 08 46c 013 10.0 points Calculate the molality of 10.5 M NH3(aq) with a density of 0.9344 g/cm3.

Nsci 14 0803

014 10.0 points Formalin is a solution of 40.0% formaldehyde (H2CO), 10.0% methyl alcohol (CH3OH), and 50.0% water by mass. Calculate the mole fraction of methyl alcohol in formalin.

Correct answer: 0.0706436.

Explanation:

In a 100 g formalin, solution the masses are

formaldehyde 40.0 g, methyl alcohol 10.0 g,

and water 50.0 g.

ntotal = nCH3OH + nH2CO + nwater

=

10 g 32 g/mol

+

40 g 30 g/mol

+

50 g 18 g/mol

= 4.42361 mol The mole fraction of methyl alcohol is

X = nCH3OH ntotal 10.0 g

=

32.0 g/mol 4.42361 mol

= 0.0706436

Correct answer: 13.8966 m.

Explanation: MW = 17.0305 g/mol

M = 10.5 M

Nsci 14 0801 015 10.0 points Toluene (C6H5CH3) is a liquid compound

Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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similar to benzene (C6H6). Calculate the mole fraction of toluene in the solution that

contains 112 g toluene and 72.0 g benzene.

Correct answer: 0.568.

Explanation:

mtoluene = 112 g

mbenzene = 72.0 g

ntoulene = (112 g toluene)

1 mol 92.14 g

= 1.21 mol

nbenzene = (72.0 g benzene)

1 mol 78.11 g

= 0.922 mol The total number of moles of all species present is

1.21 mol + 0.922 mol = 2.13 mol

The mole fraction of toluene is then

Xtoluene

=

ntoluene ntotal

=

1.21 2.13

mol mol

=

0.568

Msci 14 0802

016 10.0 points The mole fraction of a certain nonelectrolyte compound in a solution containing only that substance and water is 0.100. The molecular weight of water is 18.0 g/mol. What additional information is needed to determine the molality of the solution?

1. The density of the solute.

2. The density of the solution.

3. The molecular weight of the compound.

4. The mole fraction of water in the solution.

5. No additional information; the molality can be calculated from the information given. correct

Explanation: Here we can assume that we have 1 mol

total. (In fact, we can choose any number of moles, but the math is easier if you choose 1 mol.) If the mole fraction of the substance

is 0.100, you can then assume that you have

0.100 mol of the substance, and the remaining

0.900 mol is H2O. The molality of a solution is determined by the following formula:

m

=

mol solute kg solvent

We've already assumed that we have 0.100

mol of solute, and we can determine the kg of

H2O in the usual way:

0.900 mol H2O

18.0 g 1 mol

1 kg 1000 g

= 0.0162 kg,

and we can calculate the molality of this solu-

tion:

m

=

0.100 mol 0.0162 kg H2O

=

6.17

m

So, it is possible to determine the molality of

this solution without any additional informa-

tion.

Msci 14 0818 017 10.0 points The molecular weight of sugar is 342 and that of water is 18.01. What is the mole fraction of sugar in a 2.00 molal solution of sugar dissolved in water?

1. mole fraction = 0.0348 correct

2. mole fraction = 0.950

3. mole fraction = 0.406

4. mole fraction = 0.0360

5. mole fraction = 0.925

Explanation:

MWsugar = 342 msugar = 2.00 m

MWwater = 18.01

Molality

=

mol solute kg solvent

Molality

of

sugar

=

2.00 mol sugar 1 kg water

Mol water in 1 kg water

= (1000 g water)

1 mol water 18.01 g water

= 55.5 mol H2O

Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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The mole fraction of sugar is

Xsugar

=

mol

mol sugar sugar + mol

water

=

2.00

2.00 mol mol + 55.5

mol

= 3.48 ? 10-2

Nsci 14 0802exam 018 10.0 points 11.6 g of NaCl (58.4 g/mol) and 16.9 g of KBr (119.8 g/mol) were dissolved in 49 g of water (18.0 g/mol). What is the mole fraction of KBr in the solution?

Correct answer: 0.0312.

Explanation:

mNaCl = 11.6 g mKBr = 16.9 g mwater = 49 g

FWNaCl = 58.4 g/mol FWKBr = 119.8 g/mol FWwater = 18.0 g/mol

Correct answer: 814.286 g.

Explanation: mwater = 100 g

Xsucrose = 0.3

nwater

=

100 g 18.0 g/mol

=

5.55556

mol

Xsucrose

=

n ntotal

0.3

=

n

+

n nwater

n = 0.3 (n + nwater)

0.7 n = 0.3 nwater

n

=

0.3 nwater 0.7

=

342

m g/mol

,

so

m

=

(342

g/mol)

(0.3) 0.7

(5.55556

mol)

= 814.286 g

nKBr

=

11.6 g 119.8 g/mol

=

0.096828

mol

nNaCl

=

16.9 g 58.4 g/mol

=

0.289384

mol

nH2O

=

49 g 18.0 g/mol

=

2.72222

mol

Since ntotal = nKBr + nNaCl + nH2O,

ntotal = 0.096828 mol + 0.289384 mol + 2.72222 mol

= 3.10843 mol ,

the mole fraction is nKBr ntotal

XKBr

=

0.096828 mol 3.10843 mol

= 0.0312 mol

Nsci 14 0807

020 10.0 points A solution is made from 596 mL methanol (CH3OH of density 0.800 g/mL) and 82 mL of water (H2O of density 1.000 g/mL). Assume that the solution behaves ideally and the volumes are additive. Calculate the mole fraction of methanol in this solution.

Correct answer: 0.765848.

Explanation:

Vmethanol = 596 mL

Vwater = 82 mL

densitymethanol = 0.800 g/mL

densitywater = 1.00 g/mL

mole

fraction

CH3OH

=

nCH3OH ntotal

Nsci 14 0805

019 10.0 points How many grams of sucrose (C12H22O11) must be dissolved in 100 g water to make a solution in which the mole fraction of sucrose is 0.3?

nCH3OH = 82 mL CH3OH

0.800 g CH3OH ? 1.0 mL CH3OH

1.0 mL CH3OH ? 32.0 g CH3OH = 14.9 mol CH3OH

Version 001 ? Calculating Concentrations WKST ? vanden bout ? (51165)

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nH2O = 82 mL H2O

1.0 mL H2O 18.0 g H2O

= 4.55556 mol H2O

The mole fraction is

XCH3OH

=

14.9

14.9 mol mol + 4.55556

mol

= 0.765848

Holt da 10 rev 39

021 10.0 points Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and PN2 = 578.351 torr. What is the partial pressure of oxygen?

ntotal = 0.14 mol nH2 = 0.02 mol

Ptotal = 700 torr

XH2

=

nH2 ntotal

=

0.02 mol 0.14 mol

= 0.142857

PH2 = XH2 Ptotal = (0.142857) (700 torr) = 100 torr

Correct answer: 181.364 torr.

Explanation:

PCO2 = 0.285 torr PN2 = 578.351 torr

PT = 1 atm = 760 torr PO2 = ?

PT = PCO2 + PN2 + PO2 PO2 = PT - (PCO2 + PN2) PO2 = 760 torr - (0.285 torr + 578.351 torr) PO2 = 181.364 torr

Mlib 04 1003 022 10.0 points A 22.4 L vessel contains 0.02 mol H2 gas, 0.02 mol N gas, and 0.1 mol NH3 gas. The total pressure is 700 torr. What is the partial pressure of the H2 gas?

1. 100 torr correct

2. 7 torr

3. None of these

4. 28 torr

5. 14 torr Explanation:

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