Výpočet procenta krystalové vody v pentahydrátu thiosíranu sodného
Kolik procenta krystalové vody obsahuje:
pentahydrát thiosíranu sodného
\( \mathrm{f}\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)=\frac{6 \cdot M_{r}\left(\mathrm{H}_{2} \mathrm{O}\right)}{M_{r}\left(N a_{2} S_{2} O_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)}=\frac{108}{248,12}=0,4352 \Rightarrow \mathbf{4 3 , 5 2 \%} \mathbf{H}_{2} \mathbf{O} \)
\( \mathrm{f}\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)=\frac{4 \cdot M_{r}\left(\mathrm{H}_{2} \mathrm{O}\right)}{M_{r}\left(N a_{2} S_{2} O_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)}=\frac{72}{248,12}=0,2903 \Rightarrow \mathbf{2 9 , 0 3 \%} \mathbf{H}_{2} \mathbf{O} \)
\( \mathrm{f}\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)=\frac{5 \cdot M_{r}\left(\mathrm{H}_{2} \mathrm{O}\right)}{M_{r}\left(N a_{2} S_{2} O_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)}=\frac{90}{248,12}=0,3627 \Rightarrow \mathbf{3 6 , 2 7 \%} \mathbf{H}_{2} \mathbf{O} \)
\( \mathrm{f}\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right)=\frac{5 \cdot M_{r}\left(\mathrm{H}_{2} \mathrm{O}\right)}{M_{r}\left(N a_{2} S_{2} O_{3} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right)}=\frac{90}{230,12}=0,3911 \Rightarrow \mathbf{3 9 , 1 1 \%} \mathbf{H}_{2} \mathbf{O} \)