prepare 250.0 mL of 0.0550 mol L-1 NH4Cl(aq). Next, 10.00 mL of this solution was delivered to another 250.0 mL volumetric flask and distilled water was added to the mark. What is the concentration of the second NH4Cl(aq) solution?
To prepare 250.0 mL of 0.0550 mol L-1 NH4Cl(aq), we need to determine the amount of NH4Cl needed.
Moles of NH4Cl = Volume(L) x Concentration(mol L-1)
Moles of NH4Cl = 0.250 L x 0.0550 mol L-1
Moles of NH4Cl = 0.01375 mol
To calculate the mass of NH4Cl needed, we need to know its molar mass.
Molar mass of NH4Cl = (1 x molar mass of N) + (4 x molar mass of H) + (1 x molar mass of Cl)
Molar mass of NH4Cl = (1 x 14.01 g mol-1) + (4 x 1.01 g mol-1) + (1 x 35.45 g mol-1)
Molar mass of NH4Cl = 53.49 g mol-1
Mass of NH4Cl needed = Moles of NH4Cl x Molar mass of NH4Cl
Mass of NH4Cl needed = 0.01375 mol x 53.49 g mol-1
Mass of NH4Cl needed = 0.736 g
So, to prepare 250.0 mL of 0.0550 mol L-1 NH4Cl(aq), we need to dissolve 0.736 g of NH4Cl in enough water to make 250.0 mL of solution.
Next, 10.00 mL of this solution is transferred to another 250.0 mL volumetric flask, and distilled water is added to the mark. We can use the dilution equation to determine the concentration of the second NH4Cl(aq) solution.
C1V1 = C2V2
Where:
C1 = concentration of the first solution
V1 = volume of the first solution
C2 = concentration of the second solution
V2 = volume of the second solution
Substituting the given values, we have:
(0.0550 mol L-1)(10.00 mL) = C2(250.0 mL)
Simplifying the equation, we get:
C2 = (0.0550 mol L-1)(10.00 mL) / (250.0 mL)
C2 = 0.0022 mol L-1
Therefore, the concentration of the second NH4Cl(aq) solution is 0.0022 mol L-1.
To determine the concentration of the second NH4Cl(aq) solution, we can use the concept of dilution.
Step 1: Calculate the number of moles of NH4Cl in the original solution.
Given:
Initial solution volume (V1) = 250.0 mL = 0.250 L
Initial solution concentration (C1) = 0.0550 mol/L
Number of moles (n1) = C1 x V1 = 0.0550 mol/L x 0.250 L = 0.01375 mol
Step 2: Calculate the number of moles of NH4Cl in the final solution.
Given:
Final solution volume (V2) = 250.0 mL = 0.250 L
Final solution concentration (C2) = ? (to be determined)
Number of moles (n2) = n1 = 0.01375 mol
Step 3: Use the equation C = n/V to calculate the concentration of the final solution.
C2 = n2/V2 = 0.01375 mol / 0.250 L = 0.055 mol/L
Therefore, the concentration of the second NH4Cl(aq) solution is 0.055 mol/L.