Question

What would be the approximate weight of $1.204 \times {10^{24}}$ bromine atoms?
A.$80grams$
B.$120grams$
C.$160grams$
D.$180grams$
D.$200grams$

Answer 1

We must know about what bromine is and where it lies on the periodic table. Then we will know about atomic mass from there. We should know about how many grams of bromine contain Na atoms. Once we know that, we will find the mass of one atom. Now, we can find the weight of bromine $1.204 \times {10^{24}}$ atoms by multiplying it to the mass of one bromine atom.

Complete step by step solution:
Step1: Bromine is the chemical element whose symbol is $Br$. Its atomic number is $35$. It lies in the halogen family and it is third lightest in them. It is red-brown liquid at room temperature. It evaporates easily. Its atomic mass is $80$.
Step2: Now we can say that an $80$ gram bromine contains Na atoms(Na is Avogadro number). Because one mole of an atom contains Na atoms.
Step3: To calculate the mass of one atom we need to divide the mass with Avogadro's number and then multiply it with the given bromine atoms to find the total mass of given bromine atoms.
Total mass = $\dfrac{{Mass}}{{Na}} \times 1.204 \times {10^{24}}$ (Na is Avagadro number which is $6.022 \times {10^{23}}$ )
Total mass = $\dfrac{{80}}{{6.022 \times {{10}^{23}}}} \times 1.204 \times {10^{24}}$
Total mass = $160gm$
Hence we find that mass of $1.204 \times {10^{24}}$ bromine atoms is $160gm$.

Note: Bromine is one of the most common elements found on the earth crust. It is found in abundance at $2.4ppm$. It is found in compounds which are present in sea salt lakes. There are 2 natural isotopes but almost 24 radioactive isotopes are made in laboratories.