Final answer:
To calculate the number of methane molecules in 120 cm³ of methane at r.t.p., use Avogadro's number. The mass of 42 dm³ of chlorine at r.t.p. can be calculated using the ideal gas law and the molar mass of chlorine.
Step-by-step explanation:
In order to calculate the number of methane molecules in 120 cm³ of methane at r.t.p., we need to use Avogadro's number. Avogadro's number is 6.022 × 10^23 molecules/mol. First, we need to calculate the number of moles of methane using the ideal gas law:
n = PV/RT
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.
At r.t.p., the pressure is 1 atm and the temperature is 273 K. The ideal gas constant, R, is 0.0821 L·atm/(mol·K). Substituting these values into the equation, we get:
n = (1 atm) * (120 cm³) / (0.0821 L·atm/(mol·K) * 273 K)
n = 5.43 * 10^-3 mol
Now, we can calculate the number of molecules:
Number of molecules = n * Avogadro's number
Number of molecules = (5.43 * 10^-3 mol) * (6.022 × 10^23 molecules/mol)
Number of molecules = 3.27 x 10^21
Therefore, there are approximately 3.27 x 10^21 methane molecules in 120 cm³ of methane at r.t.p.
In order to calculate the mass of 42 dm³ of chlorine at r.t.p., we need to use the molar mass of chlorine, which is 70.906 g/mol. First, we need to calculate the number of moles of chlorine using the ideal gas law:
n = PV/RT
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature.
At r.t.p., the pressure is 1 atm and the temperature is 273 K. The ideal gas constant, R, is 0.0821 L·atm/(mol·K). Converting dm³ to L, we get:
V = 42 dm³ * (1 L / 1000 dm³) = 0.042 L
Substituting these values into the equation, we get:
n = (1 atm) * (0.042 L) / (0.0821 L·atm/(mol·K) * 273 K)
n = 2.39 * 10^-3 mol
Now, we can calculate the mass:
Mass = n * molar mass
Mass = (2.39 * 10^-3 mol) * (70.906 g/mol)
Mass = 0.170 g
Therefore, the mass of 42 dm³ of chlorine at r.t.p. is 0.170 g.