Answer to Question #165767 in Chemistry for Lwendo

Question

Calculate the amount in moles of (a) 250cm3 of sodium hydroxide solution containing 2.00mol/dm3.

(b) 200cm3 of ammonia solution having a concentration of 0.125mol/dm3

Answer 1

Solution:

(a):

Sodium hydroxide = NaOH

Molarity of NaOH = 2.00 mol/dm³ = 2.00 mol/L

Solution volume = 250 cm³ = 250 mL = 0.25 L

Molarity of NaOH = Moles of NaOH / Solution volume

Hence,

Moles of NaOH = Molarity of NaOH × Solution volume

Moles of NaOH = 2.00 mol/L × 0.25 L = 0.50 mol

Moles of NaOH = 0.50 mol

(b):

Ammonia = NH₃

Molarity of NH₃ = 0.125 mol/dm³ = 0.125 mol/L

Solution volume = 200 cm³ = 200 mL = 0.200 L

Molarity of NH₃ = Moles of NH₃ / Solution volume

Hence,

Moles of NH₃ = Molarity of NH₃ × Solution volume

Moles of NH₃ = 0.125 mol/L × 0.200 L = 0.025 mol

Moles of NH₃ = 0.025 mol

Answer:

(a): 0.50 NaOH mol

(b): 0.025 NH₃ mol