150 gm ফুটন্ত পানিতে 96 gm Pb (NO3)2 দ্রবিভূত করে একটি সম্পৃক্ত দ্রবণ তৈরী করা হল। 30°C তাপমাত্রায় Pb (NO₃)₂ এর দ্রাব্যতা 56 হলে, প্রস্তুতকৃত দ্রবণটিকে 30°C তাপমাত্রায় শীতল করা হলে কত গ্রাম দ্রব দ্রবন থেকে পৃথকীকরন হবে?

The problem is about creating a saturated solution of Pb(NO₃)₂ in water at a specific temperature and then determining how much of the solute will crystallize out when the solution is cooled.

When a solution is saturated, it means that it contains the maximum amount of solute that can dissolve in a given quantity of solvent (in this case, water) at a specific temperature. The solubility of Pb(NO₃)₂ at 30°C is given as 56 grams per 100 grams of water.

To understand the answer:

    
• We have 150 grams of water and we mixed it with 96 grams of Pb(NO₃)₂.
    
• Calculating the solubility ratio, 56 grams of Pb(NO₃)₂ can dissolve in 100 grams of water. Therefore, in 150 grams of water, the maximum amount that can dissolve is:
    
        
• (56 g / 100 g) × 150 g = 84 grams.
    
    
• Since we only mixed 96 grams of Pb(NO₃)₂, which exceeds the solubility limit, the solution is saturated, and excess solute will precipitate when cooled.
    
• When we cool the solution, it will be able to hold less solute. Thus, we need to find the excess that won’t remain dissolved at a lower temperature.
    
• The difference between the amount of solute we initially had (96 g) and the maximum amount that can stay dissolved when cooled to a lower temperature (in this case, at or below solubility limit that is around 12 grams for this situation) will crystallize out.

Thus, crystallization results in : 96 g - 84 g = 12 g.

Therefore, the answer of 12 grams crystallizing out of the solution is correct.

AI-এর ব্যাখ্যায় ভুল থাকতে পারে।