Animated Solution for Chemistry - Basic Concepts in Chemistry: The ${\text{NaNO}}_3$ weighed out to make $50\text{ mL}$ of an aqueous solution containing $70.0\text{ mg}$ ${\text{Na}}^{+}$ per mL is ……… g. (Rounded off to the nearest integer) [Given : Atomic weight in ${\text{g mol}}^{-1}$, $\text{Na}:23;\text{N}:14;\text{O}:16$].

Imagine you are standing in a pristine chemistry laboratory, tasked with preparing a very specific solution. You need exactly $50\text{ mL}$ of an aqueous solution, and it must contain precisely $70.0\text{ mg}$ of sodium ions (${\text{Na}}^{+}$) in every single milliliter. The salt you have at your disposal is solid sodium nitrate (${\text{NaNO}}_3$). The fundamental question is: how much of this white powder do you need to weigh out on the analytical balance?

This is a classic problem of stoichiometry and concentration, bridging the gap between the macroscopic world of grams and milliliters and the microscopic world of atoms and moles.

Let's break down the information given to us. We are provided with a concentration, but it's not in the standard molarity format. Instead, it's given as a mass-volume ratio: $70.0\text{ mg}$ of ${\text{Na}}^{+}$ per $1\text{ mL}$ of solution.

Our target volume is $50\text{ mL}$. To find out the total amount of sodium ions required for the entire beaker, we simply multiply the concentration by the total volume:

Notice how the milliliter units elegantly cancel out, leaving us with a total mass of $3500\text{ mg}$.

Because molar masses are conventionally expressed in grams per mole ($\text{g/mol}$), it is highly strategic to convert this mass into grams immediately to avoid any unit mismatch later. Dividing by $1000$, we get:

Entering the Mole Realm

Chemistry operates on the currency of moles, not grams. A chemical formula like ${\text{NaNO}}_3$ tells us the ratio of atoms, which directly translates to the ratio of moles. It does not tell us the ratio of masses. Therefore, our next critical step is to convert the mass of sodium ions into moles.

We use the atomic weight of sodium, which is given as $23\text{ g/mol}$.

We will leave this as a fraction for now. Evaluating it into a messy decimal too early can lead to rounding errors compounding throughout the calculation.

The Bridge

Now, we must relate the moles of the specific ion (${\text{Na}}^{+}$) to the moles of the parent compound (${\text{NaNO}}_3$). We look at the dissociation equation of the salt in water:

This equation reveals a beautiful $1:1$ stoichiometric ratio. One molecule of sodium nitrate yields exactly one sodium ion. Consequently, one mole of sodium nitrate will yield exactly one mole of sodium ions.

Because of this $1:1$ relationship, the moles of ${\text{NaNO}}_3$ we need to dissolve is exactly equal to the moles of ${\text{Na}}^{+}$ we require:

To find the physical mass we need to weigh on the balance, we must convert these moles back into grams using the molar mass of the entire compound, ${\text{NaNO}}_3$.

Let's calculate the molar mass by summing the atomic weights of its constituent atoms:

Finally, we multiply the required moles of the compound by its molar mass to find the required mass:

The problem explicitly instructs us to round off the answer to the nearest integer. Looking at $12.934$, the decimal part is greater than $0.5$, so we round up to the next whole number.

By carefully navigating through the units, converting to moles, respecting the stoichiometry, and converting back to mass, we have successfully determined the exact amount of salt needed for our solution.