Question

The concentration of chlorobenzene (C&HsCl) in water is 100 mol/m3. density is 1.00 g/cm3 The solution (a) What is the weight fraction of chlorobenzene? (b) What is the chlorobenzene concentration in PPM? (c) What is the mole fraction of chlorobenzene? (d) What is the molarity of chlorobenzene? (e) What is the molality of chlorobenzene? The concentration of chlorobenzene (C&HsCl) in air is 0.100 mol/m3 at 25 °C and 1 atm. The molecular weight of air may be taken to be 28.84 gmol. (a) What is the weight fraction of chlorobenzene? (c) What is the mole fraction of chlorobenzene? (b) What is the chlorobenzene concentration in PPM?

Answer 1

Answer:

Part 1

(a) 0.0113

(b) 11300 ppm

(c) 1.82 *10⁻³

(d) 0.100 M

(e) 0.101 m

Part 2

(a) 9.45 *10⁻³

(b) mole fraction = 2.45 *10⁻³

(c) 11.3 ppm

Explanation:

Chlorobenzene formula is C₆H₅Cl

Part 1: We are given a concentration of chlorobenzene in water of 100 mol/m³, and a density of the solution of 1.00 g/cm³.

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / mass solution

We know there are 100 moles of C₆H₅Cl per m³ of solution.

To get the mass of C₆H₅Cl we'll convert the moles to mass by using the molar mass:

Molar mass C₆H₅Cl = 6*12.011 + 5*1.00794 + 35.4527 = 112.558 g/mol

mass C₆H₅Cl = moles C₆H₅Cl * molar mass C₆H₅Cl

mass C₆H₅Cl = 100 moles * 112.558 g/mol = 11255.8 g

11255.8 g of C₆H₅Cl are in 1 m³ of solution.

Next we'll convert 1 m³ of solution to mass by using the density

mass solution = volume solution * density of solution

$mass solution = 1m^{3} *\frac{(100cm)^{3} }{ 1m^{3}} * \frac{1.00 g}{cm^{3} } = 1.00 *10^{6} g$

weight fraction C₆H₅Cl = 11256 g / 1.00 *10⁶ g = 0.0113

(b) ppm stands for "parts per million" and it is usually expressed as mg per Liter of solution

We already calculated that there are 11256 g or more exactly 11300 g of C₆H₅Cl in 1 m³ of solution, so lets convert to mg/L:

$\frac{11300 g}{1 m^{3} } * \frac{1000 mg}{1 g} * \frac{1 m^{3} }{1000 L} = 11300 mg/L$

So the solution is 11300 ppm

(c)  mole fraction = moles of C₆H₅Cl / total moles in solution

total moles = moles C₆H₅Cl + moles water

moles water = mass water / molar mass water

mass water = mass solution - mass C₆H₅Cl

moles of C₆H₅Cl = 100 moles

mass water = 1.00 *10⁶ g of solution - 11256 g = 988744 g of water

moles water = 988744 g / 18.0153 g/mol = 54884 moles water

total moles = 100 + 54884 = 54984 moles

mole fraction = 100 moles of C₆H₅Cl / 54984 moles = 1.82 *10⁻³

(d) Molarity = moles C₆H₅Cl / Liters of solution

We know the solution is 100 mol / m³ so we just have to convert the m³ to L:

$\frac{100 mol}{m^{3} } * \frac{1 m^{3}}{1000 L} = 0.100 mol / L = 0.100 M$

(e) Molality = moles C₆H₅Cl / kg water

We know that there are 100 moles per 988744 g of water, so we need to convert the grams of water to kilograms.

$Molality = \frac{100 moles}{988744 g} *\frac{1000 g}{1 kg} = 0.101 m$

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Part 2: Concentration of C₆H₅Cl in air is 0.100 mol/m³, at 25 °C and 1 atm.

Molar mass air = 28.84 g/mol

(a) weight fraction C₆H₅Cl = mass C₆H₅Cl / total mass

mass C₆H₅Cl = 0.100 mol * 112.558 g/mol = 11.26 g

total mass = mass C₆H₅Cl + mass air

mass air = moles air * molar mass air

moles air = total moles - moles C₆H₅Cl

We can calculate the total moles by using the ideal gas law:

P V = n R T

where P is pressure in atm, V is volume in L, n is the number of moles, R is the gas constant and T is temperature in Kelvin.

n = P V / R T

P = 1 atm

V = 1 m³ = 1000 L

R = 0.08206 L atm K⁻¹ mol⁻¹

T = 25 + 273.15 = 298 K

n = (1 atm * 1000 L) / (0.08206 L atm K⁻¹ mol⁻¹ * 298 K) = 40.89 moles

moles air = 40.89 - 0.100 = 40.79 moles air

mass air = 40.79 mol * 28.84 g/mol = 1176.4 g

total mass = 1176.4 g + 11.26 g = 1188 g

weight fraction = 11.26 g / 1188 g = 9.45 *10⁻³

(b) mole fraction = moles C₆H₅Cl / total moles

mole fraction = 0.100 / 40.89 = 2.45 *10⁻³

(c) ppm = mg C₆H₅Cl / Liters

We already know there are 11.26 g C₆H₅Cl in 1 m³, which is the same as 1000 L, so:

$\frac{11.26 g}{1000 L} *\frac{1000 mg}{1 g} = 11.3 mg/L$

The concentration is 11.3 ppm