Respuesta :
Answer:
A) 0.981 atm
B) 7.89 × 10⁻⁴ moles
C) 25.348 grams ≈ 25.3 grams
D) 4.315% increase error in molar mass
Explanation:
A) The partial pressure of the water vapor = 0.029 atm.
Total gas pressure in the tube = 1.01 atm
Therefore, by Dalton's law of partial pressure, the total pressure of a given mass of gas is equal to the sum of the partial pressures of the individual gases in the mixture
Hence, total pressure = partial pressure of H₂ + Partial pressure of water vapor
1.01 atm = partial pressure of H₂ + 0.029 atm.
∴ Partial pressure of H₂ = 1.01 atm - 0.029 atm = 0.981 atm
The partial pressure of the dry hydrogen gas collected in the tube = 0.981 atm
B) From the universal gas equation, we have;
PV = nRT
Therefore,
[tex]n = \frac{P \times V}{R \times T}[/tex]
Where:
n = Number of moles
P = Pressure = 0.981 atm
V = Volume = 19.6 ml
T = Temperature = 24° = (273.15 + 24) K = 297.15 K
R = Universal Gas Constant = 0.08205 L·atm/(mol·K)
Plugging in the values, we have;
[tex]n = \frac{0.981 \times 19.6 \times 10^{-3} }{0.08205 \times 297.15 } = 7.89\times 10^{-4} \ moles[/tex]
The number of moles of hydrogen gas collected = 7.89 × 10⁻⁴ moles
C) Since 1 mole of M reacts with 1 mole of H₂SO₄ in the reaction, we have
0.500 M of H₂SO₄ will react with 0.500 M of M
From;
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex] we have;
[tex]Molar \ mass = \frac{Mass}{ Number \ of \ moles} = \frac{0.0200 \, grams}{ 7.89 \times 10^{-4} \, moles} = 25.3485 \ grams[/tex]
D) The [tex]Percentage \ error = \frac{Approximate \ value - Exact \ value }{Exact \ value } \times 100 = \frac{25.3485 - 24.3}{24.3} = 4.315\%[/tex] is
Given that the actual molar mass = 24.3 g/mol.
The percentage error in the experimental molar ≈ 4.315%.
