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Reika [66]
3 years ago
10

Calculate the vapor pressure (in torr) at 298 k in a solution prepared by dissolving 15.3 g of the non-volatile non-electrolye u

rea {co(nh2)2} in 107 g of water. the vapor pressure of water at 298 k is 23.76 torr. enter your answer to 2 decimal places
Chemistry
1 answer:
ZanzabumX [31]3 years ago
7 0
In colligative properties like vapor pressure, boiling point, freezing point and osmotic pressure, only the quantity of solute may affect their values. Their identity does not affect these properties. When solute is added to a solution, the vapor pressure decreases, that's why it is called vapor pressure lowering. The equation is

ΔP = xP°, where ΔP is the difference of the vapor pressure of the solution and of the solvent. P° is the vapor pressure of the pure solvent while x is the mole fraction of solute. The molar mass of urea is 60.06 g/mol while that of the water is 18 g/mol.

x =mol urea/total mol = (15.3/60.06)/[(15.3/60.06)+(107/18)]
x = 0.0411

Then,
ΔP = xP°=0.0411*23.76 = 0.976 torr

Therefore, the vapor pressure of the solution is 23.76 - 0.976 = 22.78 Torr.
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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

6 0
3 years ago
Which gas law refers to the solubility of a gas changing with the pressure over the solution
Dmitry [639]

Answer:

the answer should be henry's law

6 0
3 years ago
Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

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3 years ago
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ExtremeBDS [4]
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6 0
3 years ago
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