1073 



0,305 

 The yield is now 100 X. 7 7. = ^! 7o • 



For the continuous process the transformations at 25° and 100*^ 

 are expressed by the equations: 



0,59 KNO J iO,36KN03 



J0,41 NaCl + 0,379 H,0 = 0,359 KNO, + 0,641 0,(U Na^M 

 2,83 H,0 ) (5,01 H,0 



{b) (c') 



and 



m,3bKN0 j /0,59KNO3 



0,641 0,64 NaCl + 0,359 NaNOa + 0,359 KCl =r 0,41 NaCl j + 

 (5,01 H,0 ] (2,83 H,0 ) 



(c') • (b) 



+ 0,359 NaCl -|- 0,379 H,0. 

 Per 1 grm. mol. of solution b only 0,359 grm. mol. of NaNO, 

 has now been converted into KNOg or 0.267 grm. mol. per 100 

 grams of solution. 



III. A solution in which KCl and NaNO^ are present in equi- 

 v^alent quantities suffers from the defect that at the isothermic 

 evaporation at 100° the saturation with KCl is already attained when 

 only a comparatively small amount of NaCl has as yet deposited. 

 The point b lies close to a. 



If to the solution is added a small excess of NaNOg much more 

 NaCl can deposit at 100°. The most favourable proportion is present 

 in a solution which passes into the liquid P^ (100°) with separation 

 of NaCl. As the composition of P^ is : 



0,38 Na, 0,62 K, 0,20 CI, 0,80 NO3, 1,81 H,0 

 this solution must contain 0.80 grm. mol. of NaNOg and 0.62 grm. mol. 

 of KCl. During the isothermic evaporation at 100° it deposits 0.42 

 mols. of solid NaCl. 



If now the solid NaCl is again removed and the liquid P^ cools 

 to 5°, KNO3 will crystallise out. The composition of the solution 

 then changes in the direction KNO3 — P^ — d. 



If no water is added, KNO3 is deposited to such an extent that 

 the solution d is attained, afterwards, along the line dP^ (5°) also 

 NaCl and finally in P^ also NaNOg. The transformation takes [)lace 

 according to the equation : 



0,62 K \ /0,12K 



0,38 Na I io,88Na 



0,80 NO3 > = 0,575 KNO3 + 0,038 NaCl + 0,010 NaNOg + 0,377 1 0,57 NO3 

 0,20 CI \ / 0,43 CI 



1,81 H2O/ U^SOHaO 



(P2 100°) (P, 5°) 



70* 



