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3 years ago

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27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a

nalysis to solve and show all work including units on every number! (5 points)

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer : The volume of the cube is, $21.88cm^3$

Solution : Given,

Density of nickel = $8.91g/cm^3$

Number of nickel atoms = $2\times 10^{24}$

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, $6.022\times 10^{23}$ atoms form 1 mole of nickel

So, $2\times 10^{24}$ atoms form $\frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321$ moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

$\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}$

$\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g$

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

$Density=\frac{Mass}{Volume}$

$8.91g/cm^3=\frac{194.94g}{Volume}$

$Volume=21.88cm^3$

Therefore, the volume of the cube is, $21.88cm^3$

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Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

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= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

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No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

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No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

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