Mass of water released =
[tex] 92.8 g CuSO_{4}.5H_{2}O [/tex]×[tex] \frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O} [/tex]
= 33.45 g [tex] H_{2}O [/tex]
Answer: The mass of water released from the given amount of copper sulfate is 33.46 grams.
Explanation:
We are given a compound having chemical formula [tex]CuSO_4.5H_2O[/tex]
The molar mass of this compound = [tex][63.55+32+(4\times 16)+5(16+(2\times 1))]=249.55g/mol[/tex]
Mass of water molecule = [tex][16+(2\times 1)]=18g/mol[/tex]
In 249.55 grams of the compound, [tex](5\times 18)g[/tex] of water molecule is present.
So, in 92.8 grams of the compound, [tex]\frac{5\times 18}{249.55}\times 92.8=33.46g[/tex] of water molecule.
Hence, the mass of water released from the given amount of copper sulfate is 33.46 grams.
