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Question: Calculate the number of moles of \( Ba{F_2} \) present in \( 41g \) of \( Ba{F_2} \) ?...

Calculate the number of moles of BaF2Ba{F_2} present in 41g41g of BaF2Ba{F_2} ?

Explanation

Solution

Hint : We must remember that the total number of atoms that are present in a mole is actually one and the same for every substance; we must be clear about this. The relation between a mole and the mass of an element is that: 1  mole  C12  atoms1\;mole\;C - 12\;atoms contain 6.023×1023  atoms6.023 \times {10^{23}}\;atoms . The steps would be to find molar mass using the molecular formula of the compound, then find the number of moles in the compound by dividing the mass by the molar mass of the compound in grams.

Complete Step By Step Answer:
Let us write down what all we know:
- For every substance, the total number of atoms inside a mole is equal.
- 1  mole  C12  atoms1\;mole\;C - 12\;atoms contain 6.023×1023  atoms6.023 \times {10^{23}}\;atoms .
Now let us write down the formula for calculating the number of moles in a given compound:
Number  of  moles=mass  of  compound(g)molar  mass(g.mol1)Number\;of\;moles = \dfrac{{mass\;of\;compound(g)}}{{molar\;mass(g.mo{l^{ - 1}})}}
But here we need to know two things: the mass of the compound and the molar mass of that compound. How do we find this?
It is given in the question that we need to find the number of moles in 41g41g of BaF2Ba{F_2} ; that is why, clearly the mass of the compound is given to be 41g41g .
To find the molar mass we can make use of our molecular formula BaF2Ba{F_2} ;
BaF2BaF_2 is made up of 1  atoms1\;atoms of Barium and 2  atoms2\;atoms of Fluorine.
\Rightarrow 1 \times 137.327\; + \;2 \times 18.9984 \\\ \Rightarrow 175.34\;g.mo{l^{ - 1}} \\\
So the molar mass of BaF2BaF_2 is 175.34  g.mol1175.34\;g.mo{l^{ - 1}} .
Now substituting the values of mass of compound and molar mass in the formula to find the number of moles, we get;
Number  of  moles=mass  of  compound(g)molar  mass(g.mol1)Number\;of\;moles = \dfrac{{mass\;of\;compound(g)}}{{molar\;mass(g.mo{l^{ - 1}})}}
= \;\dfrac{{41.0\;g}}{{175.34\;g.mo{l^{ - 1}}}} \\\
= \;0.233\;mol \\\
\therefore The number of moles of BaF2Ba{F_2} present in 41g41g of BaF2Ba{F_2} is 0.233  mol0.233\;mol .

Note :
Barium fluoride is a colorless solid. Even though its molecular weight is 175.32  gmol1175.32\;gmo{l^{ - 1}} , its exact mass is said to be 175.902  gmol1175.902\;gmo{l^{ - 1}} . It is stable when it is kept in normal conditions of storage but if it comes in contact with any mineral acid, it releases a toxic gas which is hydrogen fluoride. It is also soluble in mineral acids. It is used in the manufacturing of optical glasses, as it is a good detector of X-rays and gamma rays.