What is the molality of a solution containing 125 grams of iodine (I2) and 750 grams of CCl4? A. 1.2 m. . B. 6.57 m. . C. 0.657

Question

leonid [27]

3 years ago

14

What is the molality of a solution containing 125 grams of iodine (I2) and 750 grams of CCl4? A. 1.2 m. . B. 6.57 m. . C. 0.657

m. . D. 0.12 m.

Chemistry

1 answer:

uysha [10] · Answer 1 · 2022-03-14T00:32:17+03:00

The answer is <span>C. 0.657 m.
</span>
The molality is the number of moles of solute divided by kilograms of solvent. In a solution, solute the component which is present in lesser amount while solvent is present in greater amount. Thus, I2 is solute, and CCl4 is solvent.

So, first we need to calculate moles of I2:
Atomic mass of iodine: A(I) = 126.9
Molecular mass of iodine: Mr(I2) = 2*A(I) = 2 * 126.9 = 253.8
This means there are 253.8 g of iodine in 1 mole. So, 125 g of iodine will be in how many moles?
253.8 : 1 = 125 : x ⇒ x = 125/253.8 = 0.493 moles

Now, let's convert amount of CCl4 in kilograms (1 kg = 1000 g):
750 g = 750/1000 kg = 0.75 kg

And now, it's only left to calculate molality of the solution:
numbers of moles of solute/kilograms of solvent = 0.493 moles/0.75 kg = 0.657 m