How do we calculate the number of sodium ions, $N{a^ + }$ , in $14.2g$ of sodium sulphate? [Relative atomic masses $Na = 23,S = 32,O = 16$ ]?
Answer
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Hint: 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 $14.2g$ of sodium sulphate.
Complete step-by-step answer:sodium sulphate.First, we will write the chemical formula of Sodium Sulphate, which is $N{a_2}S{O_4}$ .
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:
$No.\,of\,moles = \dfrac{{mass\,of\,the\,compound}}{{molar\,mass\,of\,the\,\operatorname{co} compound}}$
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:
$Molar\,Mass = atomic\,mass\,in\,gram\,s = 2(23) + 32 + 4(16) = 142\,grams\,per\,mole$
Now, we have the molar mass of Sodium Sulphate, so we can find the number of Moles:
$No.\,of\,moles = \dfrac{{14.2g}}{{142gmo{l^{ - 1}}}} = 0.1mol$
Now, for every $1mol$ of $N{a_2}S{O_4}$ we have $2moles$ of $N{a^ + }$ ions.
Multiply the moles of $N{a_2}S{O_4}$ that we found just then by 2 to obtain the moles of sodium ions.
As we know, $6.022 \times {10^{23}}$ 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 $1mole$ of $N{a_2}S{O_4}$ , $6.022 \times {10^{23}}$ number of ions are present.
$\therefore $ In $0.2mole$ of $N{a_2}S{O_4}$ , $6.022 \times {10^{23}} \times 0.2\,mol$ number of ions are present, which is $1.2044 \times {10^{23}}ions$ .
Hence, $1.2044 \times {10^{23}}ions$ number of sodium ions, $N{a^ + }$ , present in $14.2g$ of
Note: A mole of a substance or a mole of particles is defined as containing exactly $6.022 \times {10^{23}}particles$ , which may be atoms, molecules, ions, or electrons. This is an experimentally obtained value.
Complete step-by-step answer:sodium sulphate.First, we will write the chemical formula of Sodium Sulphate, which is $N{a_2}S{O_4}$ .
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:
$No.\,of\,moles = \dfrac{{mass\,of\,the\,compound}}{{molar\,mass\,of\,the\,\operatorname{co} compound}}$
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:
$Molar\,Mass = atomic\,mass\,in\,gram\,s = 2(23) + 32 + 4(16) = 142\,grams\,per\,mole$
Now, we have the molar mass of Sodium Sulphate, so we can find the number of Moles:
$No.\,of\,moles = \dfrac{{14.2g}}{{142gmo{l^{ - 1}}}} = 0.1mol$
Now, for every $1mol$ of $N{a_2}S{O_4}$ we have $2moles$ of $N{a^ + }$ ions.
Multiply the moles of $N{a_2}S{O_4}$ that we found just then by 2 to obtain the moles of sodium ions.
As we know, $6.022 \times {10^{23}}$ 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 $1mole$ of $N{a_2}S{O_4}$ , $6.022 \times {10^{23}}$ number of ions are present.
$\therefore $ In $0.2mole$ of $N{a_2}S{O_4}$ , $6.022 \times {10^{23}} \times 0.2\,mol$ number of ions are present, which is $1.2044 \times {10^{23}}ions$ .
Hence, $1.2044 \times {10^{23}}ions$ number of sodium ions, $N{a^ + }$ , present in $14.2g$ of
Note: A mole of a substance or a mole of particles is defined as containing exactly $6.022 \times {10^{23}}particles$ , which may be atoms, molecules, ions, or electrons. This is an experimentally obtained value.
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