Question

How much $NaN{{O}_{3}}~$ is needed to prepare $225mL$ of a $1.55$ M solution of $NaN{{O}_{3}}~$ ?

Answer 1

Hint : Write the balanced chemical equation and then find out the ratio of the production of sodium nitrate, then use stoichiometry to calculate the number of moles produced. This means two grams of hydrogen contains atoms, and this number is also called the Avogadro’s number.

Complete Step By Step Answer:
In order to answer the question, we need to know about moles and molar mass. Now, matter is made up of atoms, and as matter has mass, then the atoms should have an individual mass. Molar mass of an element or compound is the mass which houses particles. For, example, the hydrogen molecule has a molar mass of two grams.
Number of moles of an element or a compound is the ratio of its given mass taken by the user, to its molar mass. More is the number of moles; more is the concentration of the substance. Now, let us come to the question. First, we will write the concentration;
$Concentration=\dfrac{Mass(stuff)}{MolarMass(stuff)}\times \dfrac{1}{Volume(solution)}$ and for the product,
$Number(moles)=Concentration\times Volume$
$Number(moles)=\left( 1.55M \right)\times \left( 0.225L \right)$
$\Rightarrow Number(moles)=0.348moles\left( NaN{{O}_{3}} \right)$
Thus, $0.348moles\left( NaN{{O}_{3}} \right)$ is needed equivalent to; $=\left( 0.34875\text{ }moles \right)\times \left( 84.99\text{ }g/mol \right)=29.64g.$
Therefore, $29.64g$ of $NaN{{O}_{3}}~$ is needed to prepare $225mL$ of a $1.55$ M solution of $NaN{{O}_{3}}~.$

Note :
The following method is used in order to calculate the mass to mole calculations: $Quantity(sought)=Quantity(given)\times ConversionFactor$ ; also, the number of moles is a dimensionless quantity.