Respuesta :
The calculation for the amount of water present in the given amount of hydrate is shown below,
amount water = (100 g hydrate) x (0.347 g H2O / 0.946 g hydrate)
= 36.68 g
Thus, the amount of water present in the hydrate is approximately 36.68 g.
amount water = (100 g hydrate) x (0.347 g H2O / 0.946 g hydrate)
= 36.68 g
Thus, the amount of water present in the hydrate is approximately 36.68 g.
Answer: The mass of water present in given amount of hydrate is 36.68 grams and number of moles of water are 2.04 moles.
Explanation:
We are given:
Mass of hydrate = 0.946 grams
Mass of water present = 0.347 grams
We need to calculate the mass of water present in 100 grams of hydrate. By using unitary method, we get:
In 0.946 g of hydrate, the amount of water present is 0.347 g
So, in 100 g of hydrate, the amount of water present will be = [tex]\frac{0.347g}{0.946g}\times 100g=36.68g[/tex]
Hence, the mass of water present in given amount of hydrate is 36.68 grams.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of water = 36.68 g
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of water}=\frac{36.38g}{18g/mol}=2.04mol[/tex]
Hence, the number of moles of water are 2.04 moles.