Question

What is the molar concentration of the resulting NH4OH(aq) solution when a 3.82 L volume of gaseous ammonia, NH3(g), at STP is bubbled through 100.0 mL water and reacts completely to form ammonium hydroxide, NH4OH(aq)?

Answer 1

The molar concentration of NH4OH(aq) formed when 3.82 L of NH3(g) is dissolved in 100.0 mL of water is 1.705 M.

Step-by-step explanation:

The question asks for the molar concentration of NH4OH(aq) formed when gaseous ammonia (NH3) is dissolved in water. To find the molar concentration, we need to calculate the moles of NH3 gas that reacted and then determine how many moles are present in the resulting 100.0 mL of water solution.

At standard temperature and pressure conditions (STP), 1 mole of a gas occupies 22.4 liters. First, we calculate the moles of NH3:

• Volume of NH3(g) = 3.82 L
• Moles of NH3 = Volume of NH3(g) / 22.4 L/mol

Moles of NH3 = 3.82 L / 22.4 L/mol = 0.1705 mol

Since NH3 reacts with water to form NH4OH and the reaction is a 1:1 ratio, the moles of NH3 will equal the moles of NH4OH. Next, we convert the volume of the solution from milliliters to liters:

• Volume of water = 100.0 mL = 0.1000 L

Finally, we calculate the molar concentration of NH4OH(aq) by dividing the moles of NH4OH by the volume of the solution in liters:

• Molar concentration of NH4OH(aq) = Moles of NH4OH / Volume of solution in L

Molar concentration of NH4OH(aq) = 0.1705 mol / 0.1000 L = 1.705 M