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Question: How do we calculate the number of sodium ions, \(N{a^ + }\) , in \(14.2g\) of sodium sulphate? [Rela...

How do we calculate the number of sodium ions, Na+N{a^ + } , in 14.2g14.2g of sodium sulphate? [Relative atomic masses Na=23,S=32,O=16Na = 23,S = 32,O = 16 ]?

Explanation

Solution

To solve this question, first we will calculate the molar mass of sodium sulphate, then we will find the number of moles present in it. And, finally with the help of no. of moles in sodium sulphate, we can find the number of sodium ions present in 14.2g14.2g of sodium sulphate.

Complete step-by-step answer: sodium sulphate.First, we will write the chemical formula of Sodium Sulphate, which is Na2SO4N{a_2}S{O_4} .
To find the number of sodium ions in sodium sulphate, first we have to calculate the total number of moles of Sodium Sulphate in a sample.
Here, we have the formula to calculate the number of moles of any compound:
No.ofmoles=massofthecompoundmolarmassofthecocompoundNo.\,of\,moles = \dfrac{{mass\,of\,the\,compound}}{{molar\,mass\,of\,the\,\operatorname{co} compound}}
In the above formulae, we have already given the mass of the Sodium Sulphate. But we don’t know the molar mass of the Sodium Sulphate.
So, the Molar mass of sodium sulphate is as:
MolarMass=atomicmassingrams=2(23)+32+4(16)=142gramspermoleMolar\,Mass = atomic\,mass\,in\,gram\,s = 2(23) + 32 + 4(16) = 142\,grams\,per\,mole
Now, we have the molar mass of Sodium Sulphate, so we can find the number of Moles:
No.ofmoles=14.2g142gmol1=0.1molNo.\,of\,moles = \dfrac{{14.2g}}{{142gmo{l^{ - 1}}}} = 0.1mol
Now, for every 1mol1mol of Na2SO4N{a_2}S{O_4} we have 2moles2moles of Na+N{a^ + } ions.
Multiply the moles of Na2SO4N{a_2}S{O_4} that we found just then by 2 to obtain the moles of sodium ions.
As we know, 6.022×10236.022 \times {10^{23}} is the number of particles or ions in 1mole of any substance. We use Avogadro’s Number to find the number of ions present in any compound.
So, In 1mole1mole of Na2SO4N{a_2}S{O_4} , 6.022×10236.022 \times {10^{23}} number of ions are present.
\therefore In 0.2mole0.2mole of Na2SO4N{a_2}S{O_4} , 6.022×1023×0.2mol6.022 \times {10^{23}} \times 0.2\,mol number of ions are present, which is 1.2044×1023ions1.2044 \times {10^{23}}ions .
Hence, 1.2044×1023ions1.2044 \times {10^{23}}ions number of sodium ions, Na+N{a^ + } , present in 14.2g14.2g of

Note: A mole of a substance or a mole of particles is defined as containing exactly 6.022×1023particles6.022 \times {10^{23}}particles , which may be atoms, molecules, ions, or electrons. This is an experimentally obtained value.