A 25.0-g sample of sodium hydroxide is dissolved in 400.mL of water. What is the concentration of the solution?
[NaOH] = [(25g)/(40g/mole)]/(0.400Liters) = 1.56M
In what? % w/w; % w/v, mole fraction, M, m......? If you want it in m, then
mols = grams/molar mass = ?
Then m = mols NaOH/kg solvent.
You can't do % w/v or M.
The primary reason I said M was not possible is because that definition is mols/L SOLUTION. Your problem gives 400 mL water which is not the same as 400 mL solution. 25 g NaOH + 400 mL water is not 400 mL solution but more than that. By what volume who knows. So you may calculate m but not M.
To calculate the concentration of the solution, we need to know the amount of solute (sodium hydroxide) and the volume of the solution.
Given:
Mass of sodium hydroxide (solute) = 25.0 g
Volume of solution = 400.0 mL
To find the concentration, we need to convert the volume of the solution from milliliters (mL) to liters (L). There are 1000 mL in 1 L, so:
Volume of solution = 400.0 mL ÷ 1000 = 0.400 L
Next, we need to calculate the concentration, which is defined as the amount of solute (in moles) divided by the volume of the solution (in liters). The formula is:
Concentration (in units of moles per liter) = Amount of solute (in moles) / Volume of solution (in liters)
To calculate the amount of solute in moles, we need to know the molar mass of sodium hydroxide (NaOH).
The molar mass of NaOH = (atomic mass of Na) + (atomic mass of O) + (atomic mass of H)
= (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)
≈ 39.00 g/mol
Now, we can calculate the amount of solute (NaOH) in moles:
Amount of solute (NaOH) = Mass of solute (NaOH) / Molar mass of NaOH
= 25.0 g / 39.00 g/mol
≈ 0.641 moles
Finally, we can calculate the concentration by dividing the amount of solute by the volume of the solution:
Concentration = Amount of solute / Volume of solution
= 0.641 moles / 0.400 L
≈ 1.60 moles/L
Therefore, the concentration of the sodium hydroxide solution is approximately 1.60 moles/L.