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35. How many grams of gas are present in each of the following cases?

(a) 0.100 L of CO2 at 307 torr and 26 °C

(b) 8.75 L of C2H4, at 378.3 kPa and 483 K

(c) 221 mL of Ar at 0.23 torr and -54 °C​

Respuesta :

Neetoo

Answer:

Explanation:

Given data:

A) Volume of CO₂ = 0.100 L

Pressure = 307 torr

Temperature = 26°C

Mass of CO₂ = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

26+273 = 299 K

307 torr/760 = 0.40 atm

0.40 atm × 0.100 L = n × 0.0821 atm.L/ mol.K  ×  299 K

0.04 atm.L =  n × 24.55 atm.L/ mol

n = 0.04 atm.L  / 24.55 atm.L/ mol

n = 0.002 mol

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.002 mol × 44 g/mol

Mass = 0.088 g

Given data:

b) Volume of C₂H₄ = 8.75 L

Pressure = 378.3 Kpa

Temperature = 483 K

Mass of C₂H₄  = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

378.3KPa/101 = 3.73 atm

by putting values

3.73 atm × 8.75 L = n × 0.0821 atm.L/ mol.K  ×  483 K

32.64 atm.L =  n × 39.65 atm.L/ mol

n = 32.64 atm.L / 24.55 atm.L/ mol

n = 0.82 mol

Mass of C₂H₄ :

Mass = number of moles × molar mass

Mass =  0.82 mol × 28 g/mol

Mass = 22.96 g

c)

Given data:

A) Volume of Ar = 221 mL

Pressure = 0.23 torr

Temperature = -54°C

Mass of Ar = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

-54+273 = 219 K

0.23 torr/760 = 0.0003 atm

221 mL/1000 = 0.221 L

0.0003 atm × 0.221 L = n × 0.0821 atm.L/ mol.K  ×  219 K

6.63×10⁻⁵ atm.L =  n × 17.98 atm.L/ mol

n = 6.63×10⁻⁵ atm.L  / 17.98 atm.L/ mol

n = 0.37×10⁻⁵  mol

Mass of Ar:

Mass = number of moles × molar mass

Mass = 0.37×10⁻⁵ mol × 39.9 g/mol

Mass = 14.76 ×10⁻⁵g

pstnonsonjoku

The number of moles the gas can be used to calculate the mass of the gas in each case.

Using;

a) PV = nRT

P = 307 torr or 0.4 atm

T =  26 °C + 273 = 299 K

V = 0.100 L

R = 0.082 atm LK-1mol-1

n = PV/RT

n =  0.4 atm ×  0.100 L / 0.082 atm LK-1mol-1 × 299 K

n = 0.00163 moles

Mass of CO2 =  0.00163 moles × 44 g/mol = 0.072 g

b) Again;

P = 378.3 kPa or 3.75atm

T = 483 K

V =  8.75 L

R = 0.082 atm LK-1mol-1

n = PV/RT

n = 3.75atm ×  8.75 L/0.082 atm LK-1mol-1 × 483 K

n = 0.83 moles

Mass of C2H4 =  0.83 moles × 28 g/mol = 23.24 g

c) Again;

P =  0.23 torr  or 0.0003 atm

V =  221 mL or 0.221 L

T =  -54 °C​ + 273 = 219 K

R = 0.082 atm LK-1mol-1

n = PV/RT

n =  0.0003 atm ×  0.221 L/0.082 atm LK-1mol-1 ×  219 K

n = 3.69 × 10^-6 moles

Mass of Ar  =  3.69 × 10^-6 moles × 40 g/mol

Mass of Ar  = 1.48 × 10^-4 g

Learn more: https://brainly.com/question/5624100

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