A chemist prepares a solution by dissolving 3.533 g of NaNO3 in enough water to make 200 mL of solution. What molar concentration of sodium nitrate should appear on the label? If 275 mL?

Jan 9, 2018

$c = \text{0.208 M}$ for the $\text{200 mL}$ solution and $c = \text{0.151 M}$ for the $\text{275 mL}$ solution. (not rounded for significant figures).

Explanation:

$\text{molarity" = "moles"/"liter}$

Since we have grams of ${\text{NaNO}}_{3}$, we can convert to moles of ${\text{NaNO}}_{3}$ by using the molar mass of ${\text{NaNO}}_{3}$.

${\text{3.533 g" * "1 mol"/"84.9947 g" = "0.0416 moles of NaNO}}_{3}$

So.

$\text{molarity" = ".0416 moles"/"liter}$

Now we just need to find the liters of solution. Since we have $\text{mL}$, we can convert each option directly into $\text{L}$.

$\text{200 mL"/(10^3 quad "mL") = "0.2 L of solution}$

$\text{275 mL"/(10^3 quad "mL") = "0.275 L of solution}$

Now we have all the values needed to find molarity, $c$.

$c = \text{0.0416 moles"/".2 L"= "0.208 M}$

$c = \text{0.0416 moles"/".275 L" = "0.151 M}$

So to summarize, convert grams of sodium nitrate to moles and convert milliliters of solution to liters of solution, then divide moles by liters to find molarity.

Make sure to do the math yourself, though, just to make sure I didn't make any calculator errors!