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Question: How many total atoms are there in \(0.660\,g\) of \({P_2}{O_5}\) ?...

How many total atoms are there in 0.660g0.660\,g of P2O5{P_2}{O_5} ?

Explanation

Solution

The total mass occupied by Avogadro’s number of particles I.e.,6.022×10236.022 \times {10^{23}} particles of a substance is known as its molar mass. It is generally provided in the units ofgram(gm)gram\left( {gm} \right) and is numerically equal to the molar mass of the molecule. We can easily figure this out that a mole of P2O5{P_2}{O_5} has a total of seven mole of atoms – two moles of phosphorus atoms and five moles of oxygen atoms.
In the given question, we have to find the total number of atoms in 0.660g0.660\,g of P2O5{P_2}{O_5}. We will do so by calculating moles of P2O5{P_2}{O_5} then by using mole concept and unitary method, we will solve this problem.

Complete step-by-step answer: We will first calculate the molar mass of P2O5{P_2}{O_5} =2×31+5×16=62+80=142gmol1 = 2 \times 31 + 5 \times 16 = 62 + 80 = 142g\,mo{l^{ - 1}}
We know that,
142gmol1142\,g\,mo{l^{ - 1}} or one mole of P2O5{P_2}{O_5} contains 77moles of atoms i.e., 7×NA7 \times {N_A} number of atoms =7×6.022×1023 = 7 \times 6.022 \times {10^{23}}atoms
By using unitary method, we can write
1g1g of P2O5{P_2}{O_5} will contain 7×6.022×1023142\dfrac{{7 \times 6.022 \times {{10}^{23}}}}{{142}} atoms
0.660g0.660\,g of P2O5{P_2}{O_5} will contain 7×6.022×1023142\dfrac{{7 \times 6.022 \times {{10}^{23}}}}{{142}} ×0.660 \times 0.660 atoms =1.95×1022 = 1.95 \times {10^{22}}atoms
Hence, =1.95×1022 = 1.95 \times {10^{22}} is the correct answer.

Additional information: Molar mass is the mass of 6.022×10236.022 \times {10^{23}}atoms (or one mole atoms) of an element expressed in grams.

Note: Here precaution must be taken while handling the numerical calculations as a small error in calculation could make a drastic change thus, we must try to do calculations that are precise to the very decimal points. Also, we must take note about the units of weight and molar mass in the question as this sometimes leads to grave calculative errors especially when weight is given in units other than the standard S.I. units ofgrams(gm)grams\left( {gm} \right), commonly the weight is provided in units of microgram, milligram, nanogram etc. other than the standard units of gram. Another common mistake done by students is not reading what is asked in the question properly; sometimes we are tasked with finding the number of oxygen atoms in moles and not in grams. In those cases we should only provide answers to what is asked and not to what all we can.