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Feliz [49]
3 years ago
8

A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

ature is 74.61 mmHg and its density is 0.87865 g/cm3; the vapor pressure of pure carbon tetrachloride is 91.32 mmHg and its density is 1.5940 g/cm3. If this solution is ideal, its total vapor pressure at 20°C is
Chemistry
1 answer:
finlep [7]3 years ago
7 0

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

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Explanation:

As it is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = \frac{no. of moles}{Volume in liter}

As it is given that molarity is 0.10 M and volume is 10.0 ml. As 1 ml equals 0.001 L. Therefore, 10.0 ml will also be equal to 0.01 L.

Hence, putting these values into the above formula as follows.

                  Molarity = \frac{no. of moles}{Volume in liter}

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As molar mass of KCN is equal to 65.12 g/mol. Therefore, calculate the mass of KCN as follows.

                 No. of moles = \frac{mass}{molar mass}

                                 0.001 mol = \frac{mass}{65.12 g/mol}

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Thus, we can conclude that 0.06152 grams of KCN are in 10.0 ml of a 0.10 M solution.

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