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For each solute (i) complete the following table where %i= the percent of the so

ID: 718095 • Letter: F

Question

For each solute (i) complete the following table where %i= the percent of the solute in the solutioin, Mi = molarity, mi= molality, and Xi = mole fraction of “i” .
For each solute (i) complete the following table where %i= the percent of the solute in the solutioin, Mi = molarity, mi= molality, and Xi = mole fraction of “i” .
fo achsolute completethefollow ngt blewhere %i the percent of thesolute inthese u in Me molarity, me molality, and X,- mole fraction of "You should show how you arrived at your answers show your work on a separate piece of paper s) Remember to circle your answers when you show your lute in the solutioin work Solute (olgsolute lehto volume Density %1 M Xi of Sofutionsolution of the HC,HO 5.00g 95.0g KNO CH0 54.0 1.007 /cm 1001.120 0.03764 cm 36.0 69.8 mi 100 miL 50.0% H5034.875

Explanation / Answer

Part a

For HC2H3O2

Total mass = mass of solute + Mas of H2O

= 5.00 + 95.0

= 100.0 g

Volume of solution = mass/density

= (100.0g) / (1.007 g/cm3)

= 99.3 cm3 = 99.3 mL = 0.0993 L

%i = mass of HC2H3O2 x 100 / mass of solution

= 5 x 100 / (5 + 95)

%i = 5 %

moles of HC2H3O2 = mass / molecular weight

= 5g / 60g/mol

= 0.08333 mol

Moles of water = 95g / 18g/mol = 5.27777 mol

Molarity = moles of HC2H3O2 / volume of solution

= 0.08333 mol / ( 0.0993 L)

Mi = 0.839 M

Molality = moles of HC2H3O2 / Mass of water in kg

= 0.08333 mol / ( 95 g x 1 kg/ 1000 g)

mi = 0.877 m

Total Moles = 0.08333 + 5.27777 = 5.3611 mol

Mol fraction of HC2H3O2 = moles of HC2H3O2 / total moles

= 0.08333 / 5.3611

Xi = 0.0155

Part b

For KNO3

Mass of solution = volume x density

= 1.00 L x 1000 cm3/L x 1.120 g/cm3

Mass of KNO3 + mass of water = 1120 g

w1 + w2 = 1120

w1 = 1120 - w2 ........... Eq1

Mol fraction of KNO3 = moles of KNO3 / (moles of KNO3 + moles of water)

0.03764 = n1 / (n1 + n2)

0.03764 n1 + 0.03764 n2 = n1

0.03764 n2 = 0.96236 n1

n1/n2 = 0.039112

(w1/M1) /(w2/M2) = 0.039112

(w1M2) / (w2M1) = 0.039112

(w1 x 18) / (w2 x 101.10) = 0.039112

w1/w2 = 0.21968

w1 = w2 x 0.21968 .......... Eq2

From Eq1 and eq2

w2 x 0.21968 = 1120 - w2

w2 = 918.27 g = mass of water

w1 = 1120 - 918.27 = 201.73 g = mass of KNO3

%i = mass of KNO3 x 100 / mass of solution

= 201.73 x 100 / (1120)

%i = 18.01 %

moles of KNO3 = mass / molecular weight

= 201.73g / 101.10g/mol

= 1.99535 mol

Molarity = moles of KNO3 / volume of solution

= 1.99535 mol / ( 1 L)

Mi = 1.995 M

Molality = moles of KNO3 / Mass of water in kg

= 1.99535 mol / ( 918.27 g x 1 kg/ 1000 g)

mi = 2.173 m

Part c

For C12H22O11

Density of solution = mass of solution / volume of solution

= (54 + 36) g / 69.8 mL

= 1.29 g/mL = 1.29 g/cm3

Moles of C12H22O11 = Mass / molecular weight

= 54g / 342.2965 g/mol

= 0.157757 mol

Moles of water = 36/18 = 2 mol

Mass of water in kg = 36 g x 1kg/1000g = 0.036 kg

%i = mass of C12H22O11 x 100 / mass of solution

= 54 x 100 / (54 + 36)

= 60 %

Molarity = moles of C12H22O11 / volume of solution

= 0.157757 mol / ( 69.8 mL x 1L/1000 mL)

Mi = 2.26 M

Molality = moles of C12H22O11 / Mass of water in kg

= 0.157757 mol / ( 0.036 kg)

mi = 4.38 m

Total Moles = 0.157757 + 2 = 2.157757 mol

Mol fraction of C12H22O11 = moles of C12H22O11 / total moles

= 0.157757 / 2.157757

Xi = 0.07311

Part d

For H2SO4

Mass of H2SO4 = 34.875 g

%i = 50% = mass of H2SO4 x 100 / mass of solution

0.50 x mass of solution = 34.875

Mass of solution = 69.75 g

Mass of water = 69.75 - 34.875 = 34.875 g

Volume of solution = 100 mL

Density of solution = mass of solution / volume of solution

= 69.75 g / 100 mL

= 0.6975 g/mL = 0.6975 g/cm3

Moles of H2SO4 = Mass / molecular weight

= 34.875 g / 98.079 g/mol

= 0.35558 mol

Moles of water = 34.875 g / 18 g/mol = 1.9375 mol

Mass of water in kg = 34.875 g x 1kg/1000g = 0.034875 kg

Molarity = moles of H2SO4 / volume of solution

= 0.35558 mol / ( 100 mL x 1L/1000 mL)

Mi = 3.56 M

Molality = moles of H2SO4 / Mass of water in kg

= 0.35558 mol / ( 0.034875 kg)

mi = 10.2 m

Total Moles = 0.35558 + 1.9375 = 2.29308 mol

Mol fraction of H2SO4 = moles of H2SO4 / total moles

= 0.35558 / 2.29308

Xi = 0.1551