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3 years ago

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

xtent than adding 1 mol of C6H12O6 to 1 kg of water? And why?

1 answer:

igor_vitrenko [27]3 years ago

3 0

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

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Answer:

a. I

b. I

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A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.

In the equation above

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rate = k[NO]²[O2]

where k = rate constant

From the mechanism above we see that in;

1: Rate law= k[NO]² [O2]

2: Rate law= k[N2O2][O2] [slow eq determines rate law]

3: Rate law= k[N2][O2]²

We can observe that the resembling equation is 1.

The rate of a chemical reaction is determined by the slowest step.

Rate = k[Concentration of reactants individually raised to their stoichiometric co-efficients]

In mechanism I,

Overall reaction occurs in a single step. Therefore,

rate= k[NO]²[O2]

This is this consistent with the observed rate law.

In mechanism II,

The overall reaction occurs in two steps, through the involvement of an intermediate, N2O2.

Rate of the slowest step should be the overall reaction rate.

Therefore, overall rate= k[N2O2][O2]

Again considering non-accumulation of intermediate, N2O2 in the overall reaction.

Its rate of production will be equal to its rate of decomposition.

Thus, k1[NO]²= k[N2O2][O2]

➡ [N2O2]= (k1/k). [NO]²/[O2]

Overall rate= k(k1/k).([NO]²[O2])/[O2]

=k1[NO]²

So, this is not consistent with the rate law.

Mechanism III,

the overall rate =k[NO]².

Therefore, we see that only mechanism I is is most appropriate , reasonable and consistent with the observed rate law.

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<u>Answer:</u>

<u>For a:</u> The molarity of commercial HCl solution is 12.39 M.

<u>For b:</u> The molality of commercial HCl solution is 16.79 m.

<u>For c:</u> The volume of commercial HCl solution needed is 2.42 L.

<u>Explanation:</u>

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

<u>For a:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

<u>For b:</u>

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

<u>For c:</u>

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

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