At a pressure of 9.25×10−14 atm and an ordinary temperature of 300.0 K , how many molecules are present in a volume of 1.10 cm3
1 answer:
Answer:
2.49 × 10⁶ molecules
Explanation:
Given data
- Pressure (P): 9.25 × 10⁻¹⁴ atm
- Temperature (T): 300.0 K
- Volume (V):
We can calculate the moles of gas using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 9.25 × 10⁻¹⁴ atm × 1.10 × 10⁻³ L / (0.0821 atm.L/mol.K) × 300.0 K
n = 4.13 × 10⁻¹⁸ mol
1 mole contains 6.02 × 10²³ molecules (Avogadro's number). The number of molecules in 4.13 × 10⁻¹⁸ moles is:
4.13 × 10⁻¹⁸ mol × (6.02 × 10²³ molecule/1 mol) = 2.49 × 10⁶ molecule
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Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n =
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n =
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] =
[NaA] =
Now, we will calculate the value as follows.
pH =
=
= 3.42
Thus, we can conclude that of the weak acid is 3.42.