Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm
What is the volume of 1.2 moles of water vapor at STP?
The answer is 26.9
Answer:
a and b are correct
Explanation:
This because both are aqueous solutions,therefore, identity of solvent is same that is water.
And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles
.
Hence answer is that their freezing points and Identity of the solvent shall remain the same.
Explanation:
Dissolve 93.52g of NaCl in about 400mL of distilled water, then add more water until final volume is 800mL. If starting with a solution or liquid reagent: When diluting more concentrated solutions, decide what volume(V2) and molarity (M2) the final soluble should be.
