JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
The oxygen dissolved in water exerts a partial pressure of \(20\, kPa\) in the vapour above water. The molar solubility of oxygen in water is ............ \(-\times 10^{-5}\, mol\, dm ^{-3}\) (Round off to the Nearest Integer). [Given : Henry's law constant \(= K _{ H }=8.0 \times 10^{4} kPa\) for \(O _{2}\) Density of water with dissolved oxygen \(=1.0\, kg\, dm ^{-3}\) ]
- A \(20\)
- B \(15\)
- C \(30\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\(P = K _{ H } \cdot x\) or, \(20 \times 10^{3}=\left(8 \times 10^{4} \times 10^{3}\right) \times \frac{ n _{ O _{2}}}{ n _{ O _{2}}+ n _{\text {water }}}\) or…
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