Skip to content
Exercises · 5.10

Q.34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Bihar BsebTextbookSubjectiveImportance★★★★★est
54% · 15/28 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 – Write the ideal gas equation in terms of mass and molar mass

pV=nRT=wMRT  ⟹  M=wRTpVpV = nRT = \frac{w}{M}RT \implies M = \frac{wRT}{pV}

Step 2 – Convert all data to consistent units

w=0.0625 g,V=34.05 mL=0.03405 dm3,T=546+273=819 K,p=0.1 barw = 0.0625\ \text{g}, \qquad V = 34.05\ \text{mL} = 0.03405\ \text{dm}^3, \qquad T = 546+273 = 819\ \text{K}, \qquad p = 0.1\ \text{bar}

R=0.0831 bar dm3K−1mol−1R = 0.0831\ \text{bar dm}^3\text{K}^{-1}\text{mol}^{-1}

Step 3 – Substitute

M=(0.0625)(0.0831)(819)(0.1)(0.03405)M = \frac{(0.0625)(0.0831)(819)}{(0.1)(0.03405)}

Numerator: 0.0625×0.0831×819=4.25370.0625 \times 0.0831 \times 819 = 4.2537

Denominator: 0.1×0.03405=0.0034050.1 \times 0.03405 = 0.003405 …

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.