Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
(B), because 1.0 moles would be 6.02 x 10^23 molecules. So you have half a mole.<span>
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Explanation:
1. Sedimentation and decantation cannot be used for all types of mixtures.
Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.
In decantation, gravity is used to bring the denser materials to settle at the bottom.
For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.
2. Yes, mass of an object reduces the settling time of such object in a mixture.
The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.
They can all by seperrated or replicated.