The degree of dissociation of Ca(NO3)2 in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at 100°C is 70 %. If the vapour pressure of water at 100°C is 760 mmHg, the vapour pressure of the solution is

Question

The degree of dissociation of Ca(NO3)2 in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at 100°C is 70 %. If the vapour pressure of water at 100°C is 760 mmHg, the vapour pressure of the solution is (A) 748.2 mmHg (B) 1492.6 mmHg (C) 373.2 mmHg (D) 74.03 mmHg

Tardigrade Admin · Accepted Answer

Moles of Ca(NO3)2=(7/168) present in 100.0 g of water Since dissociation is 70%, the total number of particles (7/168)× 0.7 × 3 = 0.0875=n Also, moles of solvent =(100/18)=5.55=N Applying Raoultâs law (p°-p textsolution/P°)=(n/n+N) Or (760-p/760)=(0.087/0.087+5.55) Or 760-P-760[(0.087/5.637)]=760[0.0154]=11.704 Or P=760 - 11.704=748.2 mmHg

Q. The degree of dissociation of Ca(NO3​)2​ in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at 100∘C is 70%. If the vapour pressure of water at 100∘C is 760 mmHg, the vapour pressure of the solution is

748.2 mmHg

1492.6 mmHg

373.2 mmHg

74.03 mmHg

Q. The degree of dissociation of $C a (N O_{3})_{2}$ in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at $10 0^{\circ} C$ is $70%$ . If the vapour pressure of water at $10 0^{\circ} C$ is 760 mmHg, the vapour pressure of the solution is