Question

1. silver has an atomic mass of 107.868 amu. the ag-109 isotope (108.905 amu) is 48.16%. what is the amu of the other isotope?

a) 106.905 amu
b) 106.908 amu
c) 106.903 amu
d) 106.911 amu

1. calculate the atomic mass of element \x\, if it has 2 naturally occurring isotopes with the following masses and natural abundances:

x-45 44.8776 amu 32.88%
x-47 46.9443 amu 67.12%

a) 46.26 amu
b) 45.91 amu
c) 46.34 amu
d) 46.84 amu
e) 44.99 amu

1. how many phosphorus atoms are contained in 158 kg of phosphorus?

a) 3.07 × 10²⁷ phosphorus atoms
b) 2.95 × 10²⁷ phosphorus atoms
c) 3.25 × 10²⁸ phosphorus atoms
d) 1.18 × 10²⁴ phosphorus atoms
e) 8.47 × 10²⁴ phosphorus atoms

1. calculate the mass (in g) of 2.1 × 10²⁴ atoms of w.

a) 3.9 × 10² g
b) 2.4 × 10² g
c) 3.2 × 10² g
d) 1.5 × 10² g
e) 6.4 × 10² g

Explanation:

Problem 1

Step1: Define variables

Let the atomic mass of the other isotope (Ag - 107) be \( x \) amu. The abundance of Ag - 109 is \( 48.16\%=0.4816 \), so the abundance of Ag - 107 is \( 1 - 0.4816 = 0.5184 \). The atomic mass of silver is the weighted average of its isotopes: \( 107.868=(x\times0.5184)+(108.905\times0.4816) \)

Step2: Solve for \( x \)

First, calculate \( 108.905\times0.4816 \): \( 108.905\times0.4816\approx52.45 \)
Then, rewrite the equation: \( 107.868 = 0.5184x+52.45 \)
Subtract 52.45 from both sides: \( 107.868 - 52.45=0.5184x \)
\( 55.418 = 0.5184x \)
Divide both sides by 0.5184: \( x=\frac{55.418}{0.5184}\approx106.903 \) amu

Step1: Recall the formula for atomic mass

Atomic mass \(=\sum(\text{isotope mass}\times\text{abundance}) \)

Step2: Calculate the contribution of each isotope

For X - 45: \( 44.8776\times0.3288\approx44.8776\times0.3288\approx14.76 \)
For X - 47: \( 46.9443\times0.6712\approx46.9443\times0.6712\approx31.58 \)

Step3: Sum the contributions

Atomic mass \( = 14.76+31.58 = 46.34 \) amu

Step1: Convert mass to grams

\( 158\space kg = 158\times1000 = 158000\space g \)

Step2: Calculate moles of P

Molar mass of P is \( 30.97\space g/mol \). Moles \( n=\frac{\text{mass}}{\text{molar mass}}=\frac{158000}{30.97}\approx5101.7\space mol \)

Step3: Calculate number of atoms

Using Avogadro's number \( N = n\times N_A \), where \( N_A = 6.022\times10^{23}\space atoms/mol \)
\( N = 5101.7\times6.022\times10^{23}\approx3.07\times10^{27}\space atoms \)

Answer:

C) 106.903 amu

Problem 2