PLEASE HELP!!! Just need an answer, no explanation
2 answers:
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Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.
y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
Answer:
A, C and D are correct.
Explanation:
Hello.
In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:
Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.
Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.
Best regards.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.