P = m/v
P = 0.297g/150.0ml
P = 1.98x10^-3 g/ml
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:
The limiting reactant is the Salicylic acid (C₇H₆O₃)
Explanation:
In order to find the limting reactant or the excess reactant of a chemical reaction we have to compare the moles of each reactant to the stoichiometry of the reaction; we usually make rules of three.
First of all we need to convert the mass of the reactants, to moles:
70 g / 138 g/mol = 0.507 moles of salicylic acid
80g / 102 g/mol = 0.784 moles of acetic anhydride
The reaction is: 2C₇H₆O₃ (aq) + C₄H₆O₃(aq) → 2C₉H₈O₄(aq) + H₂O(l)
Ratio is 2:1.
2 moles of salicylic acid need 1 mol of acetic anhydride to react
Then, 0.507 moles of salicylic will react with (0.507 . 1) / 2 = 0.254 moles of acetic anhydride (It's ok, I have 0.784 moles and I only need 0.254 moles, so acetic anhydride still remains, the C₄H₆O₃ is the excess reactant)
In conclussion, the limiting reactant is the Salicylic acid (C₇H₆O₃)
Let's verify: 1 mol of anhyride needs 2 moles of salicylic acid
Therefore, 0.784 moles of anhydride will react with (0.784 . 2) /1 = 1.57 moles. → We do not have enough C₇H₆O₃, we have 0.507 moles but we need 1.57.
