Respuesta :
m₀=4.052 g
m₁=1.979 g
m(H₂O)=m₀-m₁
w(H₂O)={m(H₂O)/m₀}×100=100(m₀-m₁)/m₀
w(H₂O)=100(4.052-1.979)/4.052=51.16%
51.16% H₂O
m₁=1.979 g
m(H₂O)=m₀-m₁
w(H₂O)={m(H₂O)/m₀}×100=100(m₀-m₁)/m₀
w(H₂O)=100(4.052-1.979)/4.052=51.16%
51.16% H₂O
Answer : The percentage by mass of water in the hydrate sample is, 51.16%
Explanation :
First we have to calculate the mass of water.
Mass of water = Mass of hydrated magnesium sulfate - Mass of sample after heating
Mass of water = 4.052 - 1.979 = 2.073 g
Now we have to calculate the percentage by mass of water in the sample.
Formula used :
[tex]\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of hydrated magnesium sulfate}}\times 100[/tex]
Now put all the given values in this formula, we get
[tex]\text{Mass} \%H_2O=\frac{2.073g}{4.052g}\times 100=51.16\%[/tex]
Therefore, the percentage by mass of water in the hydrate sample is, 51.16%
