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=(MolesSolute)/(Lsolution)`

Now, we find the moles of solute:

`\begin{gathered} MolesSolute=Molarity* Lsolution \\ MolesSolute=6M*400mL*(1L)/(1000mL) \\ MolesSolute=6(mol)/(L)*0.400L=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*(MolarMass,gCa(NO_3)_2)/(1molCa(NO_3)_2) \\ gCa(NO_3)_2=2.4molCa(NO_3)_2*(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