Question

1 mol = 22.4 l for a gas at stp 1 mol = 6.02 x 10²³ particles 1. how many moles of magnesium is 3.01 x 10²² atoms of magnesium? 2. how many molecules are there in 4.30 moles of glucose, c₆h₁₂o₆? 3. determine the volume, in liters, occupied by 0.030 moles of a gas at stp.

Explanation:

Question 1

Step1: Recall the mole - particle relationship

We know that \(1\space mol = 6.02\times10^{23}\space particles\) (in this case, atoms of magnesium). To find the number of moles, we use the formula \(n=\frac{N}{N_A}\), where \(n\) is the number of moles, \(N\) is the number of particles, and \(N_A = 6.02\times 10^{23}\space mol^{-1}\) (Avogadro's number).
Given \(N = 3.01\times10^{22}\space atoms\) of magnesium.

Step2: Substitute the values into the formula

\(n=\frac{3.01\times 10^{22}}{6.02\times 10^{23}}\)
We can simplify this fraction. Notice that \(3.01\times10^{22}=\frac{6.02\times 10^{23}}{20}\) (since \(6.02\times10^{23}\div20 = 3.01\times10^{22}\)). So \(\frac{3.01\times 10^{22}}{6.02\times 10^{23}}=\frac{1}{20}=0.05\space mol\)

Step1: Recall the mole - particle relationship

We know that \(1\space mol = 6.02\times10^{23}\space molecules\) (for a molecular substance like glucose). The formula to find the number of molecules \(N\) is \(N=n\times N_A\), where \(n\) is the number of moles and \(N_A = 6.02\times 10^{23}\space mol^{-1}\)
Given \(n = 4.30\space mol\) of glucose.

Step2: Substitute the values into the formula

\(N=4.30\space mol\times6.02\times 10^{23}\space molecules/mol\)
\(N = 4.30\times6.02\times 10^{23}\)
\(4.30\times6.02 = 25.886\)
So \(N=25.886\times 10^{23}=2.59\times 10^{24}\space molecules\) (rounded to three significant figures)

Step1: Recall the mole - volume relationship at STP

We know that for a gas at STP, \(1\space mol = 22.4\space L\). The formula to find the volume \(V\) of a gas is \(V=n\times V_m\), where \(n\) is the number of moles and \(V_m = 22.4\space L/mol\) (molar volume at STP)
Given \(n = 0.030\space mol\)

Step2: Substitute the values into the formula

\(V=0.030\space mol\times22.4\space L/mol\)
\(V = 0.030\times22.4\)
\(V=0.672\space L\)

Answer:

\(0.05\space mol\)

Question 2