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gtnhenbr [62]
4 years ago
5

A solution was prepared by dissolving 40.0 g of kcl in 225 g of water. part a calculate the mass percent of kcl in the solution.

express your answer with the appropriate units. 15.1 {\rm \\%} calculate the mole fraction of the ionic species kcl in the solution.
Chemistry
2 answers:
EleoNora [17]4 years ago
6 0
<h2>  part A : </h2><h2>The mass % of KCl = 17.78 %</h2>

<u>calculation </u>

 mass%= mass  of the solute/mass of the solvent x100

mass of the solute  40 g

mass of the solvent=225

% mass= 40/225 x100=17.78%


<h2>Part B</h2><h3>The mole fraction of ionic species =0.04296</h3>

the mole  fraction=moles of KCl/ moles of the solution

moles of KCl=  40 g/74.5 g/mol=0.537 moles

moles of water  = 225g/18 g/mol=12.5 moles

mole fraction is therefore= 0.537/12.5=0.04296

Alex4 years ago
3 0

Explanation:

(a)   According to the given data, mass of the solution will be calculated as follows.

          Mass of solution = Mass of KCl + Mass of water

                                      = (40 + 225) g

                                      = 265 g

Therefore, (m/m)% of KCl will be calculated as follows.

           (m/m)% of KCl = \frac{\text{mass of KCl}}{\text{mass of solution}} \times 100    

                                   = \frac{40}{265} \times 100

                                   = 15.09%

Hence, mass percent of KCl in the solution is 15.09%.

(b)    Now, we will calculate the mole fraction of ionic species present in KCl  in the solution are follows.

Molar mass of KCl is 74.55 g/mol.

                Moles of KCl = \frac{mass}{\text{molar mass}}

                                      = \frac{40 g}{74.55 g/mol}

                                      = 0.536 mol

or,                                   = 0.54 mol

As, molar mass of water is 18 g/mol. Hence, moles of water will be calculated as follows.

           Moles of H_{2}O = \frac{mass}{\text{molar mass}}

                                      = \frac{225 g}{18 g/mol}

                                      = 12.5 mol

So, total number of moles = moles of KCl + moles of H_{2}O

                                           = (0.54 + 12.5) mol

                                           = 13.04 mol

Calculate the mole fraction of KCl in the solution as follows.

              Mole fraction = \frac{\text{moles of KCl}}{\text{total moles}}

                                     = \frac{0.54 mol}{13.04 mol}

                                     = 0.041

Therefore, mole fraction of the ionic species KCl in the solution is 0.041.

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Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

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HN_3(g)\rightarrow N_2(g)+H_2(g)

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

2HN_3(g)\rightarrow 3N_2(g)+H_2(g)

As we are given:

The pressure of pure HN_3 = 3.0 atm

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From the reaction we conclude that:

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Now we have to calculate the mole fraction of N_2 and H_2

\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75

and,

\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25

Now we have to calculate the partial pressure of N_2 and H_2

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 6.0 atm

X_i = mole fraction of gas

p_{N_2}=X_{N_2}\times p_T

p_{N_2}=0.75\times 6.0atm=4.5atm

and,

p_{H_2}=X_{H_2}\times p_T

p_{H_2}=0.25\times 6.0atm=1.5atm

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

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