The concentration of a solution of ammonia, #"NH"_3# is #"1.5% m/v"#. What is
the molar concentration of a solution produced by diluting #"25.0 mL"# of this
solution with #"250 mL"# of water?

Question

Serenity Allen · Accepted Answer

#"0.080 mol L"^(-1)#

Start by calculating the number of moles of ammonia present in the #"25.0-mL"# sample of #"1.5% m/v"# ammonia solution.

As you know, a #"1.5% m/v"# solution contains #"1.5 g"# of solute, which in your case is ammonia, for every #"100 mL"# of solution.

This means that your sample will contain

#25.0 color(red)(cancel(color(black)("mL solution"))) * "1.5 g NH"_3/(100color(red)(cancel(color(black)("mL solution")))) = "0.375 g NH"_3#

Use the molar mass of ammonia to convert this to moles

#0.375 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "0.02202 moles NH"_3#

Now, you're diluting this sample by adding #"250 mL"# of water. The total volume of the diluted solution will be

#"25.0 mL + 250 mL = 275 mL"#

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the #"25.0-mL"# sample.

As you know, the molarity of a solution tells you the number of moles of solute present for every #"1 L" = 10^3# #"mL"# of solution.

In your case, the number of moles of ammonia present in #10^3# #"mL"# of this diluted solution will be equal to

#10^3 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.02202 moles H"_3/(275 color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the known composition of the diluted solution")) = "0.080 moles NH"_3#

So, if you have #0.080# moles of ammonia in #10^3color(white)(.)"mL" = "1 L"# of the diluted solution, you can say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 0.080 mol L"^(-1))))#

The answer is rounded to two sig figs.

Luna Chen · Answer

undefined To find the molar concentration of the diluted solution, first, calculate the number of moles of ammonia in the original solution, then use this to find the molar concentration of the diluted solution.

1. Calculate the number of moles of ammonia in the original solution:
Given:
- Volume of original solution (V1) = 25.0 mL
- Concentration of original solution = 1.5% m/v

Convert the volume to liters:
25.0 mL = 0.025 L

Calculate the mass of ammonia in the original solution:
Mass = Volume (L) × Concentration (g/L)
Mass = 0.025 L × (1.5 g/100 mL)
Mass = 0.025 L × (1.5 g/0.1 L)
Mass = 0.025 L × 15 g/L
Mass = 0.375 g

Now, find the number of moles of ammonia using its molar mass (M):
Molar mass of ammonia (NH3) = 14.01 g/mol (nitrogen) + 3 × 1.01 g/mol (hydrogen)
Molar mass of ammonia (NH3) = 17.03 g/mol

Number of moles = Mass (g) / Molar mass (g/mol)
Number of moles = 0.375 g / 17.03 g/mol
Number of moles ≈ 0.022 moles

2. Calculate the total volume of the diluted solution:
Total volume (V_total) = Volume of original solution + Volume of water added
Total volume (V_total) = 25.0 mL + 250 mL
Total volume (V_total) = 275.0 mL

Convert the total volume to liters:
V_total = 275.0 mL = 0.275 L

3. Calculate the molar concentration of the diluted solution:
Molar concentration (M) = Number of moles / Total volume (L)
M = 0.022 moles / 0.275 L
M ≈ 0.080 M

Therefore, the molar concentration of the diluted solution is approximately 0.080 M.

The concentration of a solution of ammonia, #"NH"_3# is #"1.5% m/v"#. What is the molar concentration of a solution produced by diluting #"25.0 mL"# of this solution with #"250 mL"# of water?