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Ainat [17]
3 years ago
12

A certain substance X has a normal boiling point of 117.8°C and a molal boiling point elevation constant =Kb·0.63°C·kgmol−1. A s

olution is prepared by dissolving some urea NH22CO in 900.g of X. This solution boils at 118.3°C. Calculate the mass of urea that was dissolved. Round your answer to 1 significant digit.
Chemistry
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

42.8 g was the mass of dissolved urea.

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

where ΔT → Boiling T° of solution - Boiling T° of pure solvent

Let's replace the data given:

118.3°C - 117.8°C = 0.63°C/m . m

0.5°C / 0.63 m/°C = m → 0.793 mol/kg

Mass of solvent → 900 g → g to kg → 900 g . 1kg/1000 = 0.9 kg

Molality . kg = moles → 0.793 mol/kg . 0.9 kg = 0.714 mol

These are the moles of urea we used. Let's determine the mass dissolved.

0.714 mol . 60 g /1mol = 42.8 g

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Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

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The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

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From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

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(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

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(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

Now put all the given values in the above equation, we get:

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

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