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Chapter 13: Problem 30

The mass of \(1 \mathrm{mol}\) of \(^{13} \mathrm{C}\) (carbon- 13 ) is $13.003 \mathrm{g}$. (a) What is the mass in \(u\) of one \(^{13} \mathrm{C}\) atom? (b) What is the mass in kilograms of one \(^{13} \mathrm{C}\) atom?

Short Answer

Expert verified
Answer: The mass of one \(^{13} \mathrm{C}\) atom is \(13.003 \ \text{u}\) in atomic mass units and \(2.1611 \times 10^{-26} \ \text{kg}\) in kilograms.

Step by step solution

01

(Find the mass in atomic mass units)

Using the given mass of 1 mol of \(^{13} \mathrm{C}\) atoms which is \(13.003 \mathrm{g}\), we will first find the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units (u). Since there are Avogadro's number (\(6.022 \times 10^{23}\) atoms) in 1 mol, we can find the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units by dividing the molar mass by Avogadro's number: $$ \text{Mass in u} = \frac{13.003 \mathrm{g/mol}}{6.022 \times 10^{23} \text{atoms/mol}} $$
02

(Calculate the mass in atomic mass units)

Now we will plug in the numbers and multiply to get our result: $$ \text{Mass in u} = \frac{13.003 \text{g/mol}}{6.022 \times 10^{23} \text{atoms/mol}} \times \frac{1 \ \text{mol}}{6.022 \times 10^{23} \ \text{atoms}} = 13.003 \ \text{u} $$ So the mass of one \(^{13} \mathrm{C}\) atom in atomic mass units is \(13.003 \ \text{u}\).
03

(Find the mass in kilograms)

Next, we will find the mass of one \(^{13} \mathrm{C}\) atom in kilograms. To do this, we will use the conversion factor between grams and atomic mass units, which is: $$ 1 \ \text{u} = 1.6605 \times 10^{-27} \ \text{kg} $$ We can multiply the mass in atomic mass units by this conversion factor to find the mass in kilograms: $$ \text{Mass in kg} = 13.003 \ \text{u} \times \frac{1.6605 \times 10^{-27} \ \text{kg}}{1 \ \text{u}} $$
04

(Calculate the mass in kilograms)

Now we will plug in the numbers and multiply to get our result: $$ \text{Mass in kg} = 13.003 \ \text{u} \times \frac{1.6605 \times 10^{-27} \ \text{kg}}{1 \ \text{u}} = 2.1611 \times 10^{-26} \ \text{kg} $$ So the mass of one \(^{13} \mathrm{C}\) atom in kilograms is \(2.1611 \times 10^{-26} \ \text{kg}\). In conclusion, the mass of one \(^{13} \mathrm{C}\) atom is \(13.003 \ \text{u}\) in atomic mass units and \(2.1611 \times 10^{-26} \ \text{kg}\) in kilograms.

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Most popular questions from this chapter

Find the rms speed in air at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm of (a) the \(\mathrm{N}_{2}\) molecules, (b) the \(\mathrm{O}_{2}\) molecules, and (c) the \(\mathrm{CO}_{2}\) molecules.
A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)
A \(10.0-\mathrm{L}\) vessel contains \(12 \mathrm{g}\) of \(\mathrm{N}_{2}\) gas at \(20^{\circ} \mathrm{C}\) (a) Estimate the nearest-neighbor distance. (b) Is the gas dilute? [Hint: Compare the nearest-neighbor distance to the diameter of an \(\left.\mathrm{N}_{2} \text { molecule, about } 0.3 \mathrm{nm} .\right]\)
A steel sphere with radius \(1.0010 \mathrm{cm}\) at \(22.0^{\circ} \mathrm{C}\) must slip through a brass ring that has an internal radius of $1.0000 \mathrm{cm}$ at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at \(22.0^{\circ} \mathrm{C},\) can just slip through?
Show that, in two gases at the same temperature, the rms speeds are inversely proportional to the square root of the molecular masses: $$ \frac{\left(v_{\mathrm{rms}}\right)_{1}}{\left(v_{\mathrm{rms}}\right)_{2}}=\sqrt{\frac{m_{2}}{m_{1}}} $$
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