Verified Solution • 2025–26 Session moderate

QJEEM25Ch604d: JEE Main 2025 Chemistry - Class 11 JEEM25Ch604d Solution: of is left after removing molecules from its ' ' mg sample. The …

2.8×10−3 mol2.8 \times 10^{-3} \mathrm{~mol} of CO2\mathrm{CO}_2 is left after removing 102110^{21} molecules from its ' xx ' mg sample. The mass of CO2\mathrm{CO}_2 taken initially is Given: NA=6.02×1023 mol−1\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}

Multiple Choice Options

  • A 98.3 mg
  • B 196.2 mg
  • C 150.4 mg
  • D 48.2 mg

Step-by-Step Verified Solution

( moles )initial =x×10−344( moles )removal =10216.02×1023( moles )left =( moles )initial −( moles )removed 2.8×10−3=x×10−344−10216.02×1023⇒x=196.2mg

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring