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3 years ago

7

Consider a solution made from a nonvolatile solute and a volatile solvent. Which statement is true?

1 answer:

34kurt3 years ago

4 0

Answer:

b) The boiling point of the solution is always greater than the boiling point of the pure solvent.

Explanation:

Hello,

In this case, when we add a nonvolatile solute to a volatile solvent which has a relatively low boiling point, we can evidence the increase of the boiling point of the resulting solution as more energy must be supplied to take the molecules from liquid to gas. This fact matches with the boiling point elevation colligative property due to the solute's addition, which states that the boiling point of the solution is always greater than the boiling point of the pure solvent, therefore, answer is b).

Best regards.

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10) How many grams are there in 1.00 x 10^24 molecules of BC13?

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194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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