With respect to #NaOH(s)#, what is the concentration of #15.5*g# mass of sodium hydroxide dissolved in a #650*mL# volume of water?

Question

Jeremiah Adams · Accepted Answer

#"Molarity"="Moles of solute"/"Volume of solution"~=0.6*mol*L^-1#

Considering the formula:

#"Molarity"=((15.50*cancelg)/(40.00*cancelg*mol^-1))/(650.0*cancel(mL)xx10^-3*L*cancel(mL^-1))=??*mol*L^-1#.

Note that this is dimensionally consistent, i.e. #1/(mol^-1)=1/(1/(mol))# #=mol# as required...........

Since we wanted an answer in #mol*L^-1# expressing #"concentration"#, and we got one, this is a good check that we performed the analysis properly. It is all too easy to divide instead of multiply or vice versa, and botch the problem entirely. We have all done it.

Henry Antrim · Answer

undefined To determine the concentration of sodium hydroxide (NaOH) in a solution, you need to calculate the molarity. First, convert the mass of NaOH from grams to moles using its molar mass, then divide by the volume of the solution in liters.

1. Calculate moles of NaOH: $ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} $
2. Convert volume to liters: $ \text{volume (L)} = \frac{\text{volume (mL)}}{1000} $
3. Calculate concentration: $ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} $

Given:
- Mass of NaOH = 15.5 g
- Volume of water = 650 mL

Molar mass of NaOH = 40.00 g/mol

1. $ \text{moles} = \frac{15.5 \text{ g}}{40.00 \text{ g/mol}} $
2. $ \text{volume (L)} = \frac{650 \text{ mL}}{1000} $
3. $ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} $

Now, calculate the concentration of NaOH.