Relative Mass Homework Question

Question

When working with gaseous substances at high temperatures and low pressures, we can use the ideal gas law N = PV/(kT) to prepare samples with specific number of particles. We just need to select specific values of pressure (P), volume (V), and temperature (T).

Imagine that the following gaseous samples of substance S1 and substance S2 are prepared. S1: P = 1 atm, T = 300 K, V = 3 L; S2: P = 1 atm, T = 300 K, V = 9 L. If the weight of sample S2 is 12 times the weight of sample S1, what is the relative mass of the particles of S2 compared to S1?

Answer Options:

a) 4

b) 3

c) 12

d) more information is needed to make a prediction

Answer 1

n_i = PV_i/RT so n₂/n₁ = V₂/V₁ = 9 L / 3 L = 3

m_i = n_iM_i (mass is moles times molar mass) So, M = m/n

M₂/M₁ =( m₂/m₁) (n₁/n₂) = 12 * 1/3 = 4

Logically, 3 times as much S2 weighs 12 times as much. Therefore, S2 has 4 times the molar mass of S1.

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