Question

In the laboratory a student was determining the formula for the hydrate of barium chloride, $Ba{Cl}_2$. The student cleaned, dried and determined the mass of the crucible and cover. The student then added a sample of the hydrate to the crucible and determined the mass. The student heated the sample in the crucible strongly for 10 minutes. The sample in the crucible and cover was cooled and the mass was determined. The mass data is provided in the table below:

Mass of crucible and cover	31.623 grams
Mass of crucible, cover, and sample prior to heating	33.632 grams
Mass of crucible, cover, and sample after heating	33.376 grams

After the heating, how many moles of water were removed from the sample?

• A. 0.0142 moles
• B. 0.0974 moles
• C. 0.112 moles
• D. 0.256 moles
• E. 1.85 moles

Answer 1

Answer: (A)

0.0142 moles

Mass of the crucible, cover and the hydrated sample before heating was $33.632grams$

Mass of the crucible, cover and the sample after heating was $33.376grams$

Clearly, some amount of water has already evaporated.

Amount of water removed is ($33.632$$-$$33.376$) grams that is $0.256grams$

Molecular weight of water is $18grams$

$18grams$ of water is equivalent to $1mole$

$1gram$of water is equivalent to $\frac{1}{18}$$mole$

$0.256grams$ of water is equivalent to $\frac{1}{18}$$\times$$0.256$$mole$ that is equal to $0.0142moles$

Option $A$ is the correct answer.