Which of the following statements is incorrect? Question 8 options: a) In genera

Question

Which of the following statements is incorrect?

Question 8 options:
  
a)   
In general, the solubility of solids in water increases with increasing temperature.

  
b)   
In general, the solubility of gases in water increases with increasing temperature.

  
c)   
In general, the solubility of gases in water increases with increasing pressure.

  
d)   
In general, the solubility of solids in water are unaffected by increasing pressure.

  
e)   
All of these statements are incorrect.

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Question 9 (1 point) Question 9 Unsaved
Which of these ions is generally insoluble in water?

Question 9 options:
  
a)   
acetate (C2H3O2–)

  
b)   
ammonium (NH4+)

  
c)   
sulfide (S2–)

  
d)   
nitrate (NO3–)

  
e)   
sodium cation (Na+)

Which solution would have a freezing point of –7.50°C?
(Kf H2O = 1.86°C/m)

Question 12 options:
  
a)   
1.34 m C2H5OH (ethanol) solution

  
b)   
2.16 m C6H12O6 (glucose) solution

  
c)   
3.50 m HOCH2CH2OH (ethylene glycol) solution

  
d)   
4.03 m C12H22O11 (sucrose) solution

  
e)   
5.86 m HOCH2CH2(OH)CH2OH (glycerol) solution

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Question 13 (1 point) Question 13 Unsaved
Which of these same solutions would have a boiling point of 103.00°C?
(Kb H2O = 0.512°C/m)

Question 13 options:
  
a)   
1.34 m C2H5OH (ethanol) solution

  
b)   
2.16 m C6H12O6 (glucose) solution

  
c)   
3.50 m HOCH2CH2OH (ethylene glycol) solution

  
d)   
4.03 m C12H22O11 (sucrose) solution

  
e)   
5.86 m HOCH2CH2(OH)CH2OH (glycerol) solution

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Question 14 (1 point) Question 14 Unsaved
How would you prepare the glucose solution (2.16 m) from the previous question?
(Mm C6H12O6 = 180.16 g/mol)

Question 14 options:
  
a)   
Add 0.720 moles of glucose to 383 grams of water.

  
b)   
Add 45.0 grams of glucose to 0.152 kg of water.

  
c)   
Add 1.45 moles of glucose to 671 grams of water.

  
d)   
Add 80.0 grams of glucose to 0.312 kg of water.

  
e)   
None of the above choices are correct.

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Question 15 (1 point) Question 15 Unsaved
What is the mass percentage of the ethylene glycol solution (3.50 m) from the previous question?
(Mm HOCH2CH2OH = 62.07 g/mol)

Question 15 options:
  
a)   
17.8 %

  
b)   
21.7 %

  
c)   
25.1 %

  
d)   
29.7 %

  
e)   
None of the above choices are correct.

Explanation / Answer

8)

we know that

solubility of gases in decreases with increase in temperature

so

the incorrect statement is b)

9) in general acetate , ammonium , sodium and nitrate ions are soluble in water

sulfide ion is insoluble in water

so

the answer is C) sulfide (S2-)

12)

we know that

depression in freezing point is given by

dTf = freezing point of water - freezing point of solution

dTf = 0 - (-7.5)

dTf = 7.5

now

we know that

dTf = i x Kf x m

7.5 = i x 1.86 x m

i x m = 4.0322

now

all the given solutions are non electrolytes

so i = vanthoff factor = 1

so

i x m = 4.0322

1 x m = 4.0322

m = 4.0322

so

the molality of the solution should be 4.0322

from the given solutions

we can see that

the solution should be sucrose

so

the answer is d) 4.03 m sucrose

13)

we know that

elevation in boiling point is given by

dTb = boiling point of solution - boiling point of water

dTb = 103 - 100

dTb = 3

now

we know that

dTb = i x Kb x m

3 = i x 0.512 x m

i x m = 5.86

now

all the given solutions are non electrolytes

so i = vanthoff factor = 1

so

i x m = 5.86

1 x m = 5.86

m = 5.86

so

the molality of the solution should be 5.86

from the given solutions

we can see that

the solution should be glycerol

so

the answer is e) 5.86 m glycerol solution

14)

we know that

molality = moles of glucose x 1000 / mass of water (g)

so

a) molality = 0.72 x 1000 / 383 = 1.88

b) moles of glucose = 45 / 180 = 0.25

molality = 0.25 x 1000 / 152 = 1.64

c) molality = 1.45 x 1000 / 671 = 2.16

d) moles of glucose = 80 / 180 = 0.4444

molality = 0.444 x 1000 / 312 = 1.424

so

the answer is c) add 1.45 moles of glucose to 671 grams of water

15)

consider 1000 g of solution

now

molality= moles of ethylene glycol x 1000 / mass of water (g)

3.5 = moles of ethylene glycol x 1000 / 1000

moles of ethylene glycol = 3.5

now

moles of ethylene glycol = 3.5 x 62.07

moles of ethylene glycol = 217.245

now

% mass = moles of ethylene glycol x 100 / mass of solution

% mass = 217.245 x 100 / 1000

% mass = 21.7

so

the answer is b) 21.7 %

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