Question

At a pressure of 9.25×10−14 atm and an ordinary temperature of 300.0 K , how many molecules are present in a volume of 1.10 cm3 ?

Answer 1

Answer:

2.49 × 10⁶ molecules

Explanation:

Given data

• Pressure (P): 9.25 × 10⁻¹⁴ atm

• Temperature (T): 300.0 K

• Volume (V): $1.10cm^{3} .\frac{1mL}{1cm^{3}} .\frac{1L}{1000mL} =1.10\times 10^{-3} L$

We can calculate the moles of gas using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 9.25 × 10⁻¹⁴ atm × 1.10 × 10⁻³ L / (0.0821 atm.L/mol.K) × 300.0 K

n = 4.13 × 10⁻¹⁸ mol

1 mole contains 6.02 × 10²³ molecules (Avogadro's number). The number of molecules in 4.13 × 10⁻¹⁸ moles is:

4.13 × 10⁻¹⁸ mol × (6.02 × 10²³ molecule/1 mol) = 2.49 × 10⁶ molecule