Answer:
The answer to your question is V2 = 1.82 l
Explanation:
Data
Volume 1 = 77 l
Pressure 1 = 18 mmHg
Volume 2 = ?
Pressure 2 = 760 mmHg
Process
Use Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (18 x 77) / 760
-Simplification
V2 = 1386 / 760
-Result
V2 = 1.82 l
Answer:

Explanation:
Hello!
In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

Thus, the percent water is:

So we plug in to obtain:

Best regards!