Question

Question 8 Calculate the volume at RTP occupied by: a) 0.16 g of oxygen ​

Answer 1

To calculate the volume of oxygen at RTP, we need to use the ideal gas law. The volume is approximately 0.4037 L.

Step-by-step explanation:

To calculate the volume of oxygen at RTP, we need to use the ideal gas law: PV = nRT. Since we have the mass of oxygen (0.16 g), we can convert it to moles using the molar mass of oxygen (32 g/mol). Next, we can use the ideal gas law to calculate the volume. The RTP (room temperature and pressure) conditions are approximately 25°C and 1 atm. So, we can substitute the values into the equation:

`V = (nRT)/(P) = \left(\frac{0.16 \, \text{g}}{32 \, \text{g/mol}}\right) \cdot \left(\frac{0.0821 \, \text{L atm}}{\text{mol K}}\right) \cdot (298 \, \text{K}) \, /\, (1 \, \text{atm}) = 0.4037 \, \text{L}`

Answer 2

0.12 dm³ or 120 cm³

Step-by-step explanation:

Hi there!

so first step find the number of moles

number of moles = mass ÷ Ar ( since o is an element and when it is alone, it is a diatomic molecule {O2} )

number of moles = mass ÷ Ar

number of moles = 0.16 ÷ ( 16 × 2 )

= 0.16 ÷ 32

= 5 × 10 ( power -3)

1 mol of O2 = 24 dm³ or 24 000 cm³

5 × 10 ( power -3) of O2 = (5 × 10 ( power -3)) × 24

= 0.12 dm³

or

= (5 × 10 ( power -3)) × 24 000

= 120 cm³