Skip to content
Exercises · 5.18

Q.2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Telangana TsbieTextbookSubjectiveImportance★★★★★est
82% · 23/28 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1 – Set up the relationship

Since both gases occupy the SAME volume VV at the SAME pressure pp (just at different temperatures), from pV=nRTpV=nRT:

n1T1=n2T2=pVR(constant, since p,V are the same for both)n_1T_1 = n_2T_2 = \frac{pV}{R} \quad (\text{constant, since } p, V \text{ are the same for both})

Step 2 – Assign the two gases

Gas 1 = unknown gas: w1=2.9 gw_1 = 2.9\ \text{g}, T1=95+273=368 KT_1 = 95+273 = 368\ \text{K}, molar mass MM (unknown)

Gas 2 = H2_2: w2=0.184 gw_2 = 0.184\ \text{g}, T2=17+273=290 KT_2 = 17+273 = 290\ \text{K}, M2=2 g mol−1M_2 = 2\ \text{g mol}^{-1}

n1=2.9M,n2=0.1842=0.092 moln_1 = \frac{2.9}{M}, \qquad n_2 = \frac{0.184}{2} = 0.092\ \text{mol} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.