Question

A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution

Answer 1

`0.745~M`

Step-by-step explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

`C_1*V_1=C_2*V_2`

Now, we have to identify the variables:

`C_1~=~14.9~M`

`V_1~=~25~mL`

`C_2~=~?`

`V_2~=~0.5~L`

Now, we have different units for the volume, so we have to do the conversion:

`0.5~L(1000~mL)/(1~L)=~500~mL`

Now we can plug the values into the equation:

`C_2=(14.9~M*25~mL)/(500~mL)=0.745~M`

I hope it helps!