Question

A 42.9 l gas sample containing 1.50 moles of oxygen at 25.0 celcius exerts a pressure of 64.0.0 torr. The volume increases to 45.0 l when additional oxygen gas is added to the sample at constant temperature and pressure. How many moles have been added? What mass of gas was added?

Answer 1

Answer:

Moles added = 0.0722 moles

Mass added = 2.3104 g

Explanation:

Given:  

Pressure = 640.0 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 640 / 760 atm = 0.8421 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25.0 + 273.15) K = 298.15 K  

Volume = 42.9 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.8421 atm × 42.9 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 1.4759 moles

Some new moles have been added and the volume has increased to 45.0 L

Using Avogadro's law

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,  

V₁ = 42.9 L

V₂ = 45.0 L

n₁ = 1.4759 moles

n₂ = ?

Using above equation as:

$\frac{42.9}{1.4759}=\frac{45.0}{n_2}$

$n_2=\frac{45.0\cdot \:1.4759}{42.9}$

n₂ = 1.5481 moles

Moles added = n₂ - n₁ = 1.5481 moles - 1.4759 moles = 0.0722 moles

Molar mass of oxygen gas = 32 g/mol

So, Mass = Moles*Molar mass = 0.0722 * 32 g = 2.3104 g

Moles added = 0.0722 moles

Mass added = 2.3104 g