How do you calculate the number of grams of solute in 250.0 mL of 0.179 M KOH?

Question

Wyatt Adkins · Accepted Answer

We use the formula.....#"Concentration"="Moles of solute"/"Volume of solution"#, and get approx. #2.5*g# #KOH#.

Thus, (i) we determine the molar quantity. Number of "Moles of KOH"="concentration"xx"volume"=0.179*mol*L^-1xx250.0*mLx10^-3*L*mL^-1=4.475x10^-2*mol. And then (ii) we multiply this molar quantity by the formula mass of #KOH#, .....#4.475xx10^-2*molxx56.11*g*mol^-1=??*g# Can you tell us #pOH# and #pH# of this solution...........?

Noah Aldrich · Answer

undefined To calculate the number of grams of solute (potassium hydroxide, KOH) in 250.0 mL of a 0.179 M (molar) solution, you can use the formula:

Number of grams of solute = Molarity × Volume × Molar mass

Given:
- Molarity (M) = 0.179 M
- Volume = 250.0 mL = 0.2500 L
- Molar mass of KOH (potassium hydroxide) = 56.11 g/mol (potassium: 39.10 g/mol, oxygen: 16.00 g/mol, hydrogen: 1.01 g/mol)

Now, plug in the values into the formula and calculate:

Number of grams of solute = 0.179 M × 0.2500 L × 56.11 g/mol

Number of grams of solute ≈ 2.507 g

Therefore, there are approximately 2.507 grams of potassium hydroxide (KOH) in 250.0 mL of a 0.179 M solution.