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A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

iris [78.8K]

To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ H2O) ---> 1

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

ρ NH3 = 0.680 g / mL

Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

Therefore the molarity of NH3 in the solution is:

Molarity = 0.881 mol / [(22.07 mL + 250 mL) * (1L / 1000 mL)

M = 3.238 mol/L = 3.24 M

8 0

3 years ago

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One of the advantages of nylon over rayon is that nylon:

daser333 [38]

- is a natural fibre
- has low elasticity
- is wrinkle free.

3 0

3 years ago

The molecular mass of octane is 114.22 g/mol. What is the mass of 22.05 mol of octane?

strojnjashka [21]

In this item, we are asked to calculate for the mass of 22.05 mole of octane given that its molecular mass is equal to 114.22 g/mol. To answer this item, we simply have to multiply the number of moles with the molecular mass. That is, (114.22 g/mol)(22.05 mol) which is equal to 2518.221 grams.

3 0

3 years ago

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

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A mixture of h2 and n2 has a density of 0.216 at 300k and 500 torr. what is the mole fraction composition of the mixture

natta225 [31]

We know that:
number of moles (n) = mass / molar mass

Now, from the general law of gases:
PV = nRT
where:
P is the pressure = 500 torr = 0.65 atm
V is the volume
n is the number of moles
R is the gas constant = 0.082
T is the temperature = 300 k
We will just rearrange this equation as follows:
P = nRT / V
Then we will substitute n with its equivalent equation mentioned at the beginning:
P = (mass x R x T) / (volume x molar mass) ......> equation I
Now, we know that:
density = mass / volume
We will substitute (mass/volume) in equation I with density as follows:
P = (density x R x T) / molar mass
Rearrange this equation to get the mass as follows:
molar mass = dRT/P = (0.216 x 0.082 x 300) / 0.65 = 8.4738 grams

From the periodic table:
molecular mass of hydrogen = 1 grams
molecular mass of nitrogen = 14 grams
Therefore:
molar mass of hydrogen = 2 x 1 = 2 grams
molar mass of nitrogen = 2 x 14 = 28 grams

We can assume that the number of moles of of each element is y.
We can thus build up the following equation:
2y + 28y = 8.4738
30y = 8.4738
y = 0.28246

Therefore:
mole fraction of hydrogen = 2 x 0.28246 = 0.56492
mole fraction of nitrogen = 28 x 0.28246 = 7.90888

4 0

4 years ago