COMEDK · Chemistry · 17. Solutions
800 mL of a \(0.5 \mathrm{M} \mathrm{HNO}_3\) at 300 K is heated to a temperature of 350 K , its volume is reduced to one half and 10.5 g of \(\mathrm{HNO}_3\) is evaporated. Assuming complete ionization, the osmotic pressure of the remaining solution is: [Given \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) ]
- A 15.5 atm
- B 16.6 atm
- C 40.2 atm
- D 33.5 atm
Answer & Solution
Correct Answer
(D) 33.5 atm
Step-by-step Solution
Detailed explanation
Initial moles of \(HNO_3\): \(n_1 = 0.5 \times 0.8 = 0.4\) mol Moles evaporated: \(n_{evap} = \dfrac{10.5}{63} = \dfrac{1}{6}\) mol Remaining moles: \(n_2 = 0.4 - \dfrac{1}{6} = \dfrac{12}{30} - \dfrac{5}{30} = \dfrac{7}{30}\) mol Final volume: \(V_2 = \dfrac{800}{2} = 400\) mL…
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