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3 years ago

What is the mole fraction of NaCl in a mixture containing 7.21 moles NaCl, 9.37 moles KCL, and 3.42

Chemistry

1 answer:

GrogVix [38]3 years ago

3 0

Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of $NaCl$ = 7.21 mole

Moles of $KCl$ = 9.37 mole

Moles of $LiCl$ = 3.42 mole

Now we have to calculate the mole fraction of $NaCl$ .

$\text{Mole fraction of }NaCl=\frac{\text{Moles of }NaCl}{\text{Moles of }NaCl+\text{Moles of }KCl+\text{Moles of }LiCl}$

Now put all the given values in this formula, we get:

$\text{Mole fraction of }NaCl=\frac{7.21}{7.21+9.37+3.42}=0.360$

Therefore, the mole fraction of NaCl in a mixture is, 0.360

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3 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

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Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

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3 years ago