The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.

advanced. .**Q1.**The amount of urea dissolved in 500 cc of water (𝐾

^{f}= 1.86℃) to produce a depression of 0.186℃ in the freezing point is

We know that,

_{f}×𝑊 / 1000×𝑘

_{f}

_{f}= 0.186℃

_{f}=1.86°, W =500 g

**Q2.**The osmotic pressure (At27℃) of an aqueous solution (200 mL) containing 6 g of a protein is 2 × 10

^{-3}atm . If R=0.080 L atm mol

^{-1}K

^{-1}, the molecular weight of protein is

𝜋𝑉 = 𝑤/𝑚𝑅𝑇 M= 𝑤𝑅𝑇 / 𝜋𝑉 Here, w=6 g, 𝜋 = 2 × 10

^{-3}𝑎𝑡𝑚, T=300 K, R=0.080 L-atm mo𝑙

^{-1}𝐾

^{-1}, V =200 mL =0.2 L M= 6 ×0.080×300 2×10

^{3}×0.2 = 3.6 × 10

^{5}

**Q3.**Assuming that sea water is a 3.50 weight per cent aqueous solution of NaCl. What is the molality of sea water?

3.50 wt% of aqueous solution of NaCl means 100 g of sea water contains 3.50 g NaCl.

**Q4.**When 25 g of a non-volatile solute is dissolved in 100 g of water, the vapour pressure is lowered by 2.25 × 10

^{-1}mm. If the vapour pressure of water at 20℃ is 17.5 mm, what is the molecular weight of the solute?

Given, Weight of non-volatile solute,

_{s}𝑝° = 𝑤×𝑀 / 𝑚×𝑊

**Q5.**Van’t hoff factor of 𝐶𝑎(𝑁𝑂

_{3})

_{2}is

**Benzoic acid in benzene exists as a dimer. So, number of molecules decreases and hence, osmotic pressure decreases**.

**Q6.**Dilute 1 L one molar H

_{2}SO

_{4}solution by 5 L water, the normality of that solution is

0.33 N

**Q7.**Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 mL solution is

𝑀 = 𝑤 × 1000 / 𝑚 × 𝑉(mL) = 75.5 × 1000 / 56 × 540 = 2.50 M

**Q8.**What is the total number of moles of 𝐻

_{2}𝑆𝑂

_{4}needed to prepare 5.0 L of a 2.0 M solution of 𝐻

_{2}𝑆𝑂

_{4}?

_{2}𝑆𝑂

^{4}= 2.0 𝑀 × 5.0 𝐿 = 10 moles

**Q9.**The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is about

Relative lowering of vapour pressure = mole fraction of solute (Raoult’s law ) 𝑝−𝑝

_{s}𝑝 = 𝑥

_{2}

**Q10.**The osmotic pressure of a solution can be accurately measured in the shortest possible time by :

It is more precise and takes minimum time.