224 mL of at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The nonvolatile solute produced is dissolved in 36 g of water. The lowering of vapour pressure of solution (assuming the

Question

224 mL of  at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The nonvolatile solute produced is dissolved in 36 g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ( mm of Hg) is  mm of Hg. The value of x is  (Integer answer)

Canvas Classes · Accepted Answer

Strategy:
 First, identify the chemical reaction between  and  to find the moles of non-volatile salt produced. Then, calculate the vapour pressure lowering using Raoult's Law.

Step 1: Determine moles of reactants
-  at 298 K, 1 atm: . (Matching JEE rounding: 0.01 mol)
- .

Step 2: Identify the product
1:1 ratio .
Moles of .

Step 3: Calculate mole fraction of solute
 mol. It dissociates (), so effective particles = 0.02 mol.
Mass of solvent (water) = 36 g  mol.

Step 4: Calculate lowering of

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