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rusak2 [61]
3 years ago
15

What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr?

The vapor pressure of pure water at 50 °C is 92.6 torr. 1
Chemistry
1 answer:
Triss [41]3 years ago
3 0

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

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Serga [27]

Answer:

-12162.47 joules (or -12000 joules when accounting for significant figures)

Explanation (btw I used 1 cal as 4.184 joules because SI units are better):

q = m c delta T

q = (70.9) (4.184) (25 - 66)  

q = (70.9) (4.184) (-41)

q = -12162.47 joules

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6 0
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A gas occupies 100 mL at 150. kPa. Find its volume at 200. kPa. You must show all your work to receive credit. Be sure to identi
PtichkaEL [24]

The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:

PV = k

where k is a constant.

We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,

P₁V₁ = P₂V₂

Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.

In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa.  Substituting these values in the previous equation and clearing V₂, we have,

P₁V₁ = P₂V₂ → V₂ = \frac{P1V1}{P2}  

→ V₂ = \frac{100 mL x 150 kPa}{200 kPa}

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Answer:

B) 16 g

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First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

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Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:

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Thus, the correct answer is option B).

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