JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(224 \,mL\) of \(SO _{2( g )}\) at \(298 K\) and \(1\, atm\) is passed through \(100 \,mL\) of \(0.1 \,M \,NaOH\) solution. The non-volatile solute produced is dissolved in \(36\, g\) of water. The lowering of vapour pressure of solution (assuming the solution is dilute)
\(\left( P _{\left( H _{2} O \right)}=24\, mm \right.\) of \(\left. Hg \right)\) is \(x \times 10^{-2} mm\) of \(Hg\) the value of \(x\) is ...... .(Integer answer)
- A \(50\)
- B \(12\)
- C \(67\)
- D \(89\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
\(SO _{2}+ NaOH \rightarrow NaHSO _{3}\) \(9.2\quad \quad \quad 10\quad \quad \quad \quad -\) \(- \quad \quad \quad 0.8 \quad \quad \quad \quad \quad 9.2\) \(\Delta P = P ^{0} \cdot X _{\text {solute }}\) \(=24 \times \frac{(1.6+18.4)}{2020}\) \(=0.2376=23.76 \times 10^{-2}\)
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