99 Percentile Qs Bank for JEE MainChemistrySolutions
A mixture of 3.0 moles of Na ₂ O and 1.5 ~mol of KO ₂ is dissolved in 1000 ~mL of water. The vapour pressure of the solution in Torr, at 100 ^ C is
Options
- A740
- B760
- C580
- D608
Correct answer
D. 608
Step-by-step solution
Na ₂ O and KO ₂ , both are ionic compounds and are completely ionised as follows : (i) Na ₂ O 2 Na ⁺+ O ²⁻(i=3) (ii) KO ₂ K ⁺+2 O ²⁻(i=3) Also, p^ -p_s p^ =i _ Na ₂ O +i _ KO ₂ and moles of H ₂ O = 1000 18 =55.5 where, p^ = atmospheric pressure p_s= vapour pressure of solution p^ =760 p^ -p_s p^ =3 3 3+3+55.5 +3 1.5 1.5+1.5+55.5 = 9 61.5 + 1.5 3 58.5 = 9 61.5 + 4.5 58.5 760-p_s 760 =0.14+0.07=0.21 0.20p_s=608