Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.
I belevieve the answer is A
