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  📘 1 Mole = Kitne Particles? Easy Explanation! "1 mole ka matlab sirf ek bada number nahi hai… ye concept JEE ke Stoichiometry ke sawalon ki chaabi hai! Agar aap mole ka matlab samajh gaye, toh chemical calculations bohot aasaan ho jaayengi." 🧪 What is 1 Mole? 1 mole = 6.022 × 10²³ particles Ye number ko Avogadro Number kehte hain. Particles ka matlab kya hota hai? 👉 Atoms, molecules, ions, electrons – jo bhi aap measure kar rahe ho, ussi ke hote hain particles. ⚖️ 1 Mole & Mass ka Connection: Har substance ka ek molar mass hota hai (grams mein). 👉 1 mole = molar mass in grams 💡 Example: Water (H₂O) ka molar mass = 18g/mol So, 18g H₂O = 1 mole = 6.022 × 10²³ molecules of water 🔁 Conversion Trick: Mole Bridge Ye ek shortcut technique hai: Mass ⇄ Moles ⇄ Number of Particles Jaise: Agar aapke paas mass ho → moles nikaalo → phir number of particles Ya particles ho → moles nikaalo → phir mass 📌 Shortcut Formula: No. of part...

  🎬 VSEPR Theory in 59 Sec — JEE Made Simple! “Bhai shape yaad karne ki tension chhodo — VSEPR theory se har molecule ka shape nikal jaata hai!” JEE ke preparation mein molecules ke shapes yaad karna ek badi headache hoti hai. But chill! Agar tumhe VSEPR Theory sahi se samajh aayi, toh tum khud se har molecule ka shape predict kar sakte ho — bina ratta lagaye! 📚 What is VSEPR Theory? VSEPR ka full form hota hai Valence Shell Electron Pair Repulsion Theory . Yeh theory ye kehti hai ki: "Electron pairs jo central atom ke around hote hain, wo ek dusre ko repel karte hain. Aur molecule apna aisa shape adopt karta hai jahan yeh repulsions minimum ho — yaani most stable shape." Iska matlab — molecular geometry ka pura funda sirf yeh samajhne mein chhupa hai ki electron pairs apas mein kitna door rehna chahte hain. 🔍 Why It Matters in JEE? Har saal JEE Mains aur Advanced dono mein VSEPR based questions aate hain. Questions directly puchhe jaate hain — "S...

  ⚛️ Bohr’s Atomic Model in 59 Seconds | Doubtify JEE Shorts 📌 Question: What explains the fixed energy levels of electrons in an atom and why do atoms emit light? Bohr’s Atomic Model gives us the first quantum explanation of atomic structure! 📸 Image: 📕 Concept & Formula: Stationary Orbits: Electrons revolve around the nucleus in fixed circular paths called orbits. No radiation is emitted in these orbits. Quantized Angular Momentum: m v r = n h 2 π mvr = \frac{nh}{2\pi} Where: m = mass , v = velocity , r = radius , n = orbit number (1, 2, 3…) , h = Planck's constant Energy Transitions: When an electron jumps between orbits: Δ E = E 2 − E 1 = h ν \Delta E = E_2 - E_1 = h\nu Energy is absorbed or emitted in the form of light/spectral lines Energy of Orbit: E n ∝ − 1 n 2 E_n \propto -\frac{1}{n^2} → Lower orbit → More negative energy → More stability 🔢 Example: An electron jumps from n = 3 to n = 2 , and emits energy. This corre...

  ⚗️ Balance the Equation: Mg + HNO₃ ⟶ Mg(NO₃)₂ + N₂O + H₂O ❓ Question: Balance the following chemical equation: Mg + HNO₃ ⟶ Mg(NO₃)₂ + N₂O + H₂O This question appears simple at first glance, but it actually requires deep understanding of redox reactions and balancing complex equations . 🎯 Key Concepts Tested: Oxidation and reduction processes Balancing redox equations using ion-electron method or oxidation number method Conservation of mass and conservation of charge Identifying oxidizing and reducing agents Understanding product formation in acidic medium (presence of HNO₃) 🧠 Step-by-Step Solution Strategy: Step 1: Identify oxidation states Let’s determine oxidation states of all elements in reactants and products: Mg (0) → Mg²⁺ in Mg(NO₃)₂ → Oxidation (Mg is losing electrons) N in HNO₃: +5 N in N₂O: +1 → Reduction (N is gaining electrons) Step 2: Write unbalanced half-reactions Oxidation (Mg): Mg → Mg²⁺ + 2e⁻ Reduction (N): 2...

🔥 Combustion Analysis – Percentage of Oxygen in an Organic Compound | JEE Chemistry | Doubtify JEE

🧪 Question: The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6:1. If one molecule of the above compound contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO₂ and H₂O, then what is the empirical formula of the compound? 📌 Why This Question is Important: This question combines two highly testable concepts in JEE Chemistry – combustion analysis and empirical formula derivation . You’re not only working with mass percent ratio , but also applying oxygen balancing logic in a combustion reaction. Such questions are frequently seen in organic chemistry chapters , particularly under Basic Principles of Organic Chemistry and Stoichiometry . Mastering this helps you crack more complex analysis-based questions in JEE Mains and even Advanced . 💡 Concept Used: To find the empirical formula , we first convert the mass percent ratio of C and H into moles , assuming 100 g of substance. Then we apply a stoichiometric appr...

🧮 Question: 100 grams of propane is completely reacted with 1000 grams of oxygen. The mole fraction of carbon dioxide in the resulting mixture is X × 10⁻². The value of x is: 🖼️ Question Image:

     🧲 Solutions  | Chemistry  |  Doubtify JEE 🧮 Question: The density of the NaOH solution is 1.2 g/cm³. What is the molality of the solution? 🖼️ Question Image: ⚗️ Concept Overview: This is a fundamental problem from the Solutions chapter of Physical Chemistry. It tests your understanding of molality , density , and the method of converting between different concentration units. In many JEE Main and Advanced problems, students get confused between molality (mol/kg) and molarity (mol/L) . This problem helps to eliminate that confusion and improve conceptual clarity. 🧠 Step-by-Step Explanation: Let’s assume we are working with 1000 cm³ = 1 L of NaOH solution. Density = 1.2 g/cm³ So, total mass of 1 L solution = 1.2   g/cm³ × 1000   cm³ = 1200   g 1.2 \, \text{g/cm³} \times 1000 \, \text{cm³} = 1200 \, \text{g} Let’s say the solution is x% NaOH by mass (you may be given the percentage or molarity in the full version of the question – for no...

🧪 Redox Reactions | Chemistry |  JEE Mains 📌 Question: Balance the equation by ion-electron method in basic medium: Zn + NO 3 − → Zn 2 + + NH 4 +​ This is a classic redox equation where zinc is oxidized and nitrate is reduced. The balancing must be done using the ion-electron method under basic medium conditions . 🖼️ Question Image: 🧠 Solution (Image): 🎥 Video Solution: 🔍 Why this Question is Important: This question is frequently asked in JEE Mains and Advanced , especially in the chapter Redox Reactions . It tests key skills like: Identifying oxidation and reduction half-reactions Using the ion-electron method properly in basic medium Applying concepts of oxidation states, electrons, and balancing H₂O/OH⁻ ions Understanding this concept is crucial for solving more advanced redox problems in competitive exams. 📩 Have a Doubt? Drop us a mail at  doubtifyqueries@gmail.com  or DM us on  Instagram . Stay consistent. Watch – Pause – Solve – Repeat. ...