Answer:
Twice as much.
Explanation:
That's because the freezing point depression depends on the total number of solute particles.
C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)
0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).
That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.
Answer:
reacts to form carbondioxide
<u>Answer:</u> The temperature of the system is 273 K
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of carbon dioxide = 1 lb = 453.6 g (Conversion factor: 1 lb = 453.6 g)
Molar mass of carbon dioxide = 44 g/mol
Putting values in above equation, we get:

To calculate the temperature of gas, we use the equation given by ideal gas equation:
PV = nRT
where,
P = Pressure of carbon dioxide = 200 psia = 13.6 atm (Conversion factor: 1 psia = 0.068 atm)
V = Volume of carbon dioxide =
(Conversion factor:
)
n = number of moles of carbon dioxide = 10.31 mol
R = Gas constant = 
T = temperature of the system = ?
Putting values in above equation, we get:

Hence, the temperature of the system is 273 K