Answer:
Molarity = 0.809 M
mole fraction = 0.047
Explanation:
The complete question is
Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Solution -
Solution for molarity:
1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.
1)
Mass of 1.09 mole of acetone
= 1.09 mol x 58.0794 g/mol = 63.306 g
Density of acetone = 0.788 g/cm3
Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3
For ethanol
1000 g divided by 0.789 g/cm3 = 1267.427 cm3
Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3 = 1.347 L
a) Molarity:
1.09 mol / 1.347 L = 0.809 M
Mole Fraction
a) moles of ethanol:
1000 g / 46.0684 g/mol = 21.71 mol
b) moles of acetone:
1.09 / (1.09 + 21.71) = 0.047
Answer: The values become more negative
Explanation: I just took the quiz and got it correct :)
Answer:
496 g of Fe₂O₃.
Explanation:
The balanced equation for the reaction is given below:
4Fe + 3O₂ —> 2Fe₂O₃
From the balanced equation above,
4 moles of Fe reacted to produce 2 moles of Fe₂O₃.
Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃
Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:
Mole of Fe₂O₃ = 3.1 moles
Molar mass of Fe₂O₃ = (56 × 2) + (3×16)
= 112 + 48
= 160 g/mol
Mass of Fe₂O₃ =?
Mass = mole × molar mass
Mass of Fe₂O₃ = 3.1 × 160
Mass of Fe₂O₃ = 496 g
Therefore, 496 g of Fe₂O₃ were produced from the reaction.
A colloid is a heterogeneous mixture in which the dispersed particles are intermediate in size between those of a solution and a suspension. The particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas.
Answer:
8.1%
Explanation:
<h3><u><em>mass of glucose/ total mass of solution x 100</em></u></h3>
So first 265mg into grams = 0.265 g
0.265 + 3 = 3.265
mass of glucose= 0.265
total mass of solution= 3.265
0.265/3.265 x 100 = 8.1% (rounded to 2 s.f)
