Question

question 24
1 pts
how many particles are in 78.9g of mgbr₂?
a. 4.75x10²³ particles
b. 4.75x10⁻²³ particles
c. 2.58x10⁻²³ particles
d. 2.58x10²³ particles

Explanation:

Step1: Calculate molar mass of MgBr₂

The molar mass of Mg (magnesium) is approximately 24.31 g/mol, and the molar mass of Br (bromine) is approximately 79.90 g/mol. Since there are 2 Br atoms in MgBr₂, the molar mass of MgBr₂ is $M = 24.31+2\times79.90=24.31 + 159.8=184.11$ g/mol.

Step2: Calculate number of moles

The number of moles $n$ of MgBr₂ is calculated using the formula $n=\frac{m}{M}$, where $m = 78.9$ g and $M=184.11$ g/mol. So $n=\frac{78.9}{184.11}\approx0.4285$ mol.

Step3: Calculate number of particles

According to Avogadro's number ($N_A = 6.022\times 10^{23}$ particles/mol), the number of particles $N$ is $N=n\times N_A$. Substituting $n = 0.4285$ mol and $N_A=6.022\times 10^{23}$ particles/mol, we get $N=0.4285\times6.022\times 10^{23}\approx2.58\times 10^{23}$ particles.

Answer:

D. $2.58\times 10^{23}$ particles