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miv72 [106K]
3 years ago
8

Estimate the freezing and normal boiling points of 0.25 m aqueous solutions of nh4no3 nicl3 al2so43

Chemistry
1 answer:
maks197457 [2]3 years ago
7 0
The items are answered below and are numbered separately for each compound. 

The freezing point of impure solution is calculated through the equation,
     Tf = Tfw - (Kf)(m)

where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.

1. NH4NO3
    Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C

2. NiCl3 
    Tf = 0°C - (1.86°C/m)(0.25 M)(4) = -1.86°C

3. Al2(SO4)3
      Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C

For boiling points, 
    Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.

1. NH4NO3
     Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C

2. NiCl3
     Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C

3. Al2(SO4)3
    Tb = 100°C + (0.51°C/m)(0.25 M)(5) = 100.6375°C
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monitta

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Explanation:

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3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

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Answer:

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Hydrogen has an electronegativity of 2.20 and oxygen has 3.44. That means that oxygen attracts the electrons more strongly than hydrogen does (second statement).

As consequence, the electrons in the covalent bond H - O of water are not shared equally (fourth statement): the electron density will  be higher around the O atoms.

Of course, this discards the statement telling that hydrogen atom attracts electrons much more strongly than the oxygen atom, and the statement telling that hydrogen and oxigen have same electronegativity.

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