Question

6. You have 400 mL of a 6 M solution of Ca(NO3)2. How many grams are present?

Answer 1

We are given the liters and molarity of the solution. To find grams we must take into account the definition of molarity. Molarity is defined as:

$Molarity=\frac{MolesSolute}{Lsolution}$

Now, we find the moles of solute:

$\begin{gathered} MolesSolute=Molarity\times Lsolution \\ MolesSolute=6M\times400mL\times\frac{1L}{1000mL} \\ MolesSolute=6\frac{mol}{L}\times0.400L=2.4mol \end{gathered}$

Now, to convert these moles to grams we must multiply the moles by the molar mass of Ca(NO3)2. Molar mass Ca(NO3)2 is 164.10g/mol.

So, the grams will be:

$\begin{gathered} gCa(NO_3)_2=givenmolCa(NO_3)_2\times\frac{MolarMass,gCa(NO_3)_2}{1molCa(NO_3)_2} \\ gCa(NO_3)_2=2.4molCa(NO_3)_2\times\frac{164.10gCa(NO_3)_2}{1molCa(NO_3)_2} \\ gCa(NO_3)_2=394gCa(NO_3)_2 \end{gathered}$

Answer: There are present 394grams of Ca(NO3)2