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ratelena [41]
3 years ago
8

Which aqueous solution has the lowest boiling point?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer : The correct option is, (a) 1.25 M C_6H_{12}O_6

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

where,

\DeltaT_b = change in boiling point

k_b = boiling point constant

m = molality

i = Van't Hoff factor

As per question, we conclude that the molality of the given solution are same. So, the boiling point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) C_6H_{12}O_6 is a non-electrolyte solute that means they retain their molecularity, an not undergo association or dissociation.

So, Van't Hoff factor = 1

(b) The dissociation of KNO_3 will be,

KNO_3\rightarrow K^++NO_3^-

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(C) The dissociation of KNO_3 will be,

AlCl_3\rightarrow Al^[3+}+3Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The boiling point depends only on the Van't Hoff factor.That means lower the Van't Hoff factor, lower will be the boiling point and vice-versa.

Hence, the correct option is, (a) 1.25 M C_6H_{12}O_6

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Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
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1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

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