Question

5) Calculate the number of grams of solute that has to be used to make the following solution:152.9 mL of 0.452 M NaNO3

Answer 1

The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of the solution expressed in liters:

`\text{Molarity (M)=}\frac{mole\text{s of solute}}{litres\text{ of solution}}=(moles)/(L).`

We have to convert 152.9 mL to liters (L). Remember that 1 liter equals 1000 mL:

`152.9mL\cdot(1L)/(1000mL)=0.1529\text{ L.}`

And we have to solve for 'moles' from the formula of molarity:

`\text{moles = Molarity }\cdot\text{ L.}`

And finally, we replace the given molarity (0.452 M) and the volume of solution which is 0.1529 L:

`\begin{gathered} \text{moles = 0.452 M}\cdot0.1529L, \\ \text{moles}=0.0691\text{ moles.} \end{gathered}`

Now, that we have the number of moles of NaNO3, we have to calculate its mass in grams. We have to use the molar mass of NaNO3 which is 85 g/mol (you can calculate the molar mass using the periodic table). The conversion from 0.0691 moles of NaNO3 to grams would be:

`0.0691molesNaNO_3\cdot\frac{85\text{ g }NaNO_3}{1\text{ mol }NaNO_3}=5.87\text{ g }NaNO_3.`

The answer is that we have 5.87 grams of NaNO3 in 152.9 mL of 0.452 M of NaNO3.