Question

What is the mass of $NaCl$ required to prepare $0.5$ liters of a $2.5$ molar solution of $NaCl$ ?

Answer 1

To solve this type of question firstly we need to calculate the number of moles $NaCl$ of at present. The mole is a unit of measurement for an amount of the subsequent in system unit that too not standard but International System. A mole of the substance is the mole of particles which are defined as containing exactly.

Complete solution:
The molecular weight of $NaCl$ can be defined as $58.44\text{ }\dfrac{g}{mol}$ Here its cleared that one mole of $NaCl$ weights around $58.44\text{ }g$
Also as per the given values, a $2.5\text{ }M$ solution is taken here $2.5$ moles per liter is just Molarity and its just the number of moles per liter.
Therefore, we know that $0.5\text{ }L$ would contain $1.25$ mol right now we need to calculate for molecular weight $1.25\text{ }\times \text{ }58.44\text{ }g\text{ }=\text{ }73\text{ }g$
The following sets can be written as
$M\text{ }=\text{ }moles/vol\text{ }\ldots \ldots \ldots \ldots \left( i \right)$
Here $M$ is for molarity.
$vol\text{ }=\text{ }volume\text{ }\left( in\text{ }liters \right)$
$moles\text{ }=\text{ }\dfrac{grams}{MW}\text{ }\ldots \ldots \ldots \ldots \left( ii \right)$
Here $g$ is the weight of a compound and $MW$ is molecular weight.
Substituting the value of moles in equation $\left( i \right)$ we get;
$M\text{ }=\text{ }\dfrac{\left( \dfrac{g}{MW} \right)}{vol}\text{ }\ldots \ldots \ldots \ldots \left( iii \right)$
Here we need to find the value of $g$
Therefore, rearranging the equation $\left( iii \right)$ we get;
$g\text{ }=\text{ }M\times ~MW\times ~vol\text{ }\ldots \ldots \ldots \ldots \left( iv \right)$
Substituting the acquired values in equation $\left( iv \right)$
$g=2.5\dfrac{mol}{L}\times 0.5\text{ }L\times 58.44\dfrac{g}{mol}$
$g\text{ }=\text{ }73g$
Therefore $73g$ mass of $NaCl$ required to prepare $0.5$ liters of a $2.5$ molar solution of $NaCl$.

Note: Note that while calculating mass of compound we need acquire molecular weight of Compound without molar mass we can't get further also the important formula for determining mass $M\text{ }=\text{ }\dfrac{moles}{vol}$ and likewise $moles\text{ }=\text{ }\dfrac{grams}{MW}$ and after substituting we acquire the mass of $NaCl$ or required compound.