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dolphi86 [110]
3 years ago
6

I need help on homework

Chemistry
1 answer:
My name is Ann [436]3 years ago
3 0
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A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that
UkoKoshka [18]

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>

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If the actual yielsd of a reaction is 50 g and the theoretical yield is 60 g. What is the percent yield.
balandron [24]
M₁=50 g
m₀=60 g

w=100m₁/m₀

w=100*50/60=83.3%
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