Question

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Question 31 (1 point) A solution was made by adding 87 g of ethanol, C2HOH, to 742 of water. How much would this lower the freezing point? The freezing point depression constant. Kt. of water is 1.86 °C-m-1 and the molar mass of ethanol is 46.08 g-mol-1. (Don't use the degree sign in your answer.) Your Answer: Answer units Question 32 (1 point) What is the pH of a 0.023 M aqueous acetic acid solution? Kg - 1.8 x 10-5 Your Answer: Answer Question 33 (1 point) What is the pH of a 0.033 M solution of aqueous Sr(OH)2? Your Answer:

Answer 1

31)

Mass of Ethanol ( solute) = 87 g

mass of water ( solvent) = 742 g

Molality of the solution , m =

m solutemass(g) MW 1000 masso sof solvent (9)

  

m 879 1000 * 46g/mol 742g

= 2.55 m

Depression in the freezing point is colligative property.

Ethanol is non eletrolyte , unionizable.( so vant Hoff factor = 1)

Depression in the freezin gpoint ,

AT, = freezingpointconstant * molality

  

AT, = 1.86°C/m 1.86°C/m * 2.55m

=

4.743°C

The freezin gpoint is lowered by 4.74

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32)

Acetic acid is weak acid.So it's degree of dissociation is to be determined to find the concentration of H^+ ion concentration.

CH3COOH + CH3C00- + H+

Let Degree of dissociation of acetic acid = x  

Degree of dissociation is the fraction of 1 mole of acetic acid that dissociates into its ions,

	CH3COOH	CH3C00	H +
Initial	0.023M	0	0
Change	-0.023x	+0.023x	+0.023x
Equilibrium	0.023-0.023x	0.023x	0.023x

So = 0.023 x

H +

But,

Ka = сHзсоо [H+] |CHзсоон

Κα = 0.02310.0230 0.023(1 – 1)

Κα= (0.023r)2 0.023(1 – 1)

Since Ka is very small the demominaor is assumed to be equalto 0.023 , [1-x =1 ( Approximation)]

So,

1.8* 10-5 = 0.023.2

1.8 * 10-5 T 0.023

= 0.02797

So hydrogen ion concetration , [H^+] = 0.023 X 0.02797

= 0.0006434M

= 6.43 * 10^-4 M

pH is the negative logarthim of hydrogen ion conectration

pH = -log[+]

= -log( 6.43 * 10^-4 M)

= 3.2

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33)

Concentration of Sr(OH)2 = 0.033M

Sr(OH)2 Sr2+ + 2OH

So ,

lmolSr(OH)2 = 2molOH

So,

  

OH = 2 * 0.033М

= 0.066 M

pH = 14-pOH

pOH = -log[OH^-]

= -log ( 0.066)

= 1.18

So , pH = 14 - 1.18

= 12.82

pH of the given solution = 12.82

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30 )

The given compounds are

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Methylene chloride - Tetrahedral

H and Cl both have differet electronegativities. Chlorine is more electronegative than hydrogen . So the molecule possess net dipole moment. Polar molecule.

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dichlorodifluoromethane - tetrahedral

F and Cl both have differet electronegativities. Chlorine is les electronegative than Fluorine . So the molecule possess net dipole moment. Polar molecule.

ANswer = Both are polar.

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29)

Nitrate ion

Formal charge on any atom , F = valence electrons of that atom ( v) - [ non bonding electrons of that atom + number of bonds]

OR

Bondedelectrons F=V- 2

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valence electrons on nitrogen = 5

NUmber of bonds = 4 ( or bonded electrons = 8/2)

non bonded electrons on nitrogen atom = 0

Formal charge on Nitrogen atom in nitrate ion , F = valence electrons of that atom ( v) - [ non bonding electrons of that atom + number of bonds]

F = 5 - 4

= + 1

Formal charge on Nitrogen atom = +1