496 



Messrs. T. and A. Gray and J. J. Dobbie. [Apr. 24, 



tion with a crystal glass analysed by Berthier, to which he assigns 

 the formula — 



2(K 2 0.3Si0 2 ) + 3(Pb0.3Si0 2 ). 



Specimen V. 



2Pb0.3K 2 0.1S'a 2 0.24Si0 2 ; 



or, 



3{ (Pb0.2Si0 3 ) + 2(K 2 0.5Si0 2 ) } + {Pb0.2Si0 2 ) + 2 (Na 2 0.5Si0 2 ) } 



Found. Calculated. 



SiO a V. 64-21 64-55 



PbO 20-50 20-0 



K 2 G 1231 12-66 



Na 3 2-97 2*77 



Specimen VI. 



This flask has obviously the same composition as V. 



Specimen X. 



Eliminating the lime and trace of magnesia, we have as expressing 

 the composition of this glass — 



4Pb0.5K 2 0.5Na 2 0.48Si0 2 ; 



or, (2(Pb0.2Si0 2 ) + 5(K 2 0.4Si0 2 ) }, 



mixed with (2(Pb0.2Si0 3 ) + 5(Na 3 0.4Si0 3 )} 



Found. Calculated. 



Si0 2 63-09 63-29 



PbO 19-62 19-59 



K 2 9-77 10-34 



Na,0 7-5 6-8 



99-98 



100-02 



or, 



mixed with 



Specimen XX. 

 PbO.K 2 ONa 3 O.10SiO 2 ; 

 { (PbO.2 Si0 2 ) + 2 (K 2 0.4Si0 2 ) } 

 {(Pb0.2Si0 2 ) +2(Na 2 0.4Si0 3 )} 



Found. 



62-5 



Si0 2 



PbO 22-76 



K 2 

 Na,0 



8-33 

 6-3 



Calculated. 



61-28 

 22-77 

 9-62 

 6-33 



99-89 



100-00 



