<span>1. Fill a beaker or graduated cylinder with enough water to completely immerse the sphere in. 2. Record the baseline initial measurement. 3. Drop the sphere in. 4 <span>Record final measurement.</span></span>
Molar mass CH4 = 16.0 g/mol
* number of moles:
932.3 / 16 => 58.26875 moles
T = 136.2 K
V = 0.560 L
P = ?
R = 0.082
Use the clapeyron equation:
P x V = n x R x T
P x 0.560 = 58.26875 x 0.082 x 136.2
P x 0.560 = 650.76
P = 650.76 / 0.560
P = 1162.07 atm
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
Answer:
There are 2500000 milligrams in 2.5 kg
