Calculate the weight proportion of H2O in Na2S2O2.5H2O?

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Calculate the weight proportion of H2O in Na2S2O2.5H2O?

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Answer 1 · 2017-10-21T14:09:36

Are you sure you don't mean $sodium thiosulfate pentahydrate$ $"sodium thiosulfate pentahydrate"$ , i.e. $N a_{2} S_{2} O_{3}$ $Na_2S_2O_3$ ? I make $percentage water$ $"percentage water"$ approx. $60 %$ $60%$ .

Explanation:

We can calculate the percentage of water BY mass in $N a_{2} S_{2} O_{3} \cdot 5 H_{2} O$ $Na_2S_2O_3*5H_2O$ ....

i.e. $= \frac{Mass of water}{Mass of hydrate} = \frac{5 \times 18.01 \cdot g \cdot m o l^{- 1}}{158.11 \cdot g \cdot m o l^{- 1}} \times 100 %$ $="Mass of water"/"Mass of hydrate"=(5xx18.01*g*mol^-1)/(158.11 *g*mol^-1)xx100%$

$= ? ? %$ $=??%$

Just in regard to $S_{2} O_{3}^{2 -}$ $S_2O_3^(2-)$ , I have always liked to imagine this as $S O_{4}^{2 -}$ $SO_4^(2-)$ where ONE of the oxides has been replaced by sulfide ion, the which assumes the SAME oxidation number as oxide, and thus we gots $S (V I +)$ $S(VI+)$ and $S (- I I)$ $S(-II)$ , an AVERAGE oxidation state of $S (+ I I)$ $S(+II)$ .

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Calculate the weight proportion of H2O in Na2S2O2.5H2O?

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