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Vadim26 [7]
2 years ago
14

Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling

point of the solvent
Pure dissolved in 0.284 ° C. Determine the Kb value of solvent B. Set the molecular weight of A = 106.
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
8 0

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

T_{solution}-T{solvent}=K_bm_solute

Considering that the difference in the temperature is 0.284°C, and the given molality by:

m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m

Now, solving for K_b, we get:

K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m

Best regards.

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Initial                      3.4 atm           1.3 atm                         0 atm

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