What volume of a 0.580-m solution of cacl2 contains 1.28 g solute?
1 answer:
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Answer:
T = 246 K
Explanation:
Given that,
Number of moles, n = 0.750 mol
The volume of the cylinder, V = 6850 mL = 6.85 L
Pressure of the gas, P = 2.21 atm
We need to find the temperature of the gas stored in the cylinder. We know that,
PV= nRT
Where
R is gas constant
T is temperature
So,
or
T = 246 K
So, the temperature of the gas is equal to 246 K.
