BREAKING NEWS

Breaking News - Career Updates
📢 Latest Job & Exam Updates — CareerInformationPortal.in 🔥 Punjab PSPCL JE Electrical Admit Card 2026 Out | July 31, 2026 🔗 Check Full Details     |     🏛️ Patna High Court Assistant Recruitment 2026: Online Form Started | Last Date: 27th August 2026 🔗 Apply / Check Details     |     🔬 UPSSSC Forensic Science Laboratory Recruitment 2026: Online Form | Last Date: 17th August 2026 🔗 Apply / Check Details     |     🏦 PNB Bank Local Bank Officer (LBO) Recruitment 2026: Online Form | Last Date: 9th August 2026 🔗 Apply / Check Details     |     📚 RSSB CET 12th Level Online Form 2026 Started: Apply Now | Last Date: 23rd July 2026 🔗 Apply / Check Details     |     ✨ Stay Updated – Bookmark for Daily Sarkari Naukri Alerts. 🙏 🔗 https://www.careerinformationportal.in/

Website Search

SELF STUDY

SELF STUDY
SELF STUDY

CAREER JOB

JOB CAREER HUB
For ALL GOVT& PRIVATE JOBS

APNA CAREER

Tuesday, August 11, 2026

Calculate the mass of 3.011×10 23 molecules of CO 2 ​ .

📘 Step-by-Step Solution

Step 1: Given Data

Number of CO₂ molecules = 3.011 × 10²³

We have to calculate the mass of these CO₂ molecules.

Step 2: Find the number of moles

We know that Avogadro's number is:

NA = 6.022 × 10²³ molecules mol⁻¹

Use:

Number of moles = Number of molecules ÷ NA

Therefore:

n = (3.011 × 10²³) ÷ (6.022 × 10²³)

n = 0.5 mol

🎬 Molecule → Mole Animation

C O O 0.5 mol CO₂

3.011 × 10²³ molecules = 0.5 mole of CO₂

Step 3: Calculate molar mass of CO₂

Atomic masses:

C = 12 g mol⁻¹

O = 16 g mol⁻¹

Therefore:

Molar Mass of CO₂

= 12 + (2 × 16)

= 12 + 32

= 44 g mol⁻¹

Step 4: Calculate the mass

Use the formula:

Mass = Number of moles × Molar Mass

Substitute the values:

Mass = 0.5 × 44

= 22 g

🎬 Final Calculation

3.011 × 10²³ molecules 0.5 mol 44 g mol⁻¹ 22 g

✅ Final Answer

Mass of CO₂ = 22 g

Therefore, 3.011 × 10²³ molecules of CO₂ have a mass of 22 g.

🎯 Quick Formula

Moles = Molecules ÷ 6.022 × 10²³

Mass = Moles × Molar Mass

CO₂ Molar Mass = 44 g mol⁻¹

Mass = 22 g