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lesya692 [45]
3 years ago
7

Choose the solvent below that would show the greatest freezing point lowering when used to make a 0.20 m nonelectrolyte solution

. diethyl ether, kf = 1.79°c/m benzene, kf = 5.12°c/m chloroform, kf = 4.70°c/m ethanol, kf = 1.99°c/m carbon tetrachloride, kf = 29.9°c/m
Chemistry
1 answer:
tatiyna3 years ago
4 0

Answer : Carbon tetrachloride, k_f=29.9^oC/m will show the greatest freezing point lowering.

Explanation :

For non-electrolyte solution, the formula used for lowering in freezing point is,

\Delta T_f=k_f\times m

where,

\Delta T_f = lowering in freezing point

k_f = molal depression constant

m = molality

As per question, the molality is same for all the non-electrolyte solution. So, the lowering in freezing point is depend on the k_f only.

That means the higher the value of k_f, the higher will be the freezing point lowering.

From the given non-electrolyte solutions, the value of k_f of carbon tetrachloride is higher than the other solutions.

Therefore, Carbon tetrachloride, k_f=29.9^oC/m will show the greatest freezing point lowering.

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Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0
3 years ago
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