Question

The average concentration of $N{a^ + }$ ion in the human body is $3{\text{ to }}4g{\text{ per litre}}$. The approximate molarity of $N{a^ + }$ ion in the body is:
A.) $0.15M$
B.) $0.2M$
C.) $0.25M$r
D.) $0.3M$

Answer 1

Hint: To solve this question first find out the number of moles of the sodium ion as the mass is given then secondly find out the molarity that is given as division of number of moles to the volume of the solution.

Complete step by step answer:
Molarity is defined as the moles of a solute in per liter of a solution. Molarity is also known as the molar concentration of a solution. Basically, the molar concentration is the measure of the concentration of solute in a solution. Molarity or molar concentration tells us how concentrated a substance is. The formula or molar concentration or molarity can be represented as:
$M = \dfrac{n}{V}$ $ - (1)$
Where, $M = $ Molarity of the solute or molar concentration of the solute in the solution
$n = $ number of moles of solute
$V = $ Volume of solution.
To find the molarity, first we need a number of moles of solute. As we know that the number of moles ($n$) can be given as:
$n = \dfrac{W}{m}$
Where, $W = $ given mass ( that is the mass of $N{a^ + }$ ion given $3 - 4g$ )
$m = $ molecular mass
As we know that the molecular mass of sodium is $23g$ and we will suppose the average given mass as $3.5g$. So, now the number of moles can be given as:
$
  n = \dfrac{{3.5}}{{23}} \\
  n = 0.15{\text{ moles}} \\
 $
Now, to find molarity the volume is given as $1litre$ (for $3 - 4g$ of solute) and the number of moles is $0.15$. Therefore, the equation $ - (1)$ becomes
$
  M = \dfrac{{0.15}}{1} \\
  M = 0.15M \\
 $
Hence the molarity of sodium ion is $0.15M$.

Therefore, option A.) is the correct answer.

Note:
In such questions, we may confuse for finding the molarity because to find molarity we need the volume of solution but remember that $3{\text{ to }}4g{\text{ per litre}}$ is given that is for one $litre$ of solution we have $3 - 4g$ of the ions. So, we will take volume as one $litre$.