Kat070
contestada

A 2.000-g sample of a hydrated copper sulfate is heated, and the waters of hydration are removed. If the mass of the remaining salt is 1.278 g, what was the percentage of water in the original hydrate?

Use the formula %water=(mass of water/mass of hydrate)•100

-13.0%
-36.1%
-56.5%
-72.2%

Respuesta :

kenmyna
  The %   of water in the original  hydrate  is  36.1%

      Calculation
    this is calculated by use of the  formula   %   water   mass of water/ mass  of the  hydrate  x100

 mass  of  water = mass of hydrate -  mass  of  unhydrate

= 2.00 g - 1.278  = 0.722  grams
mass of hydrate =2.00 g

 % water is therefore = 0.722g/2.00  x100=  36.1 %

B or 2nd option: 36.1%

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