How many milligrams (mg) would you use to make 15 ml of a 250 millimolar (MM) KCl solution? Answer mg 6. How many milligrams (mg) would you use to make 75 ml of a 800 micromolar (UM) KCl solution? Answer mg
The correct answer and explanation is:
Answers:
- For 15 ml of a 250 mM KCl solution:
279.56 mg - For 75 ml of an 800 µM KCl solution:
4.47 mg
Explanation
To prepare chemical solutions accurately, understanding molar concentration calculations is essential. The two given problems demonstrate converting molar concentrations into mass measurements required for making laboratory solutions.
1. Preparing 15 ml of a 250 mM KCl solution:
First, convert volume to liters: 15 ml=151000=0.015 L15\,ml = \frac{15}{1000} = 0.015\,L
Next, convert concentration from millimolar (mM) to molar (M): 250 mM=2501000=0.25 mol/L250\,mM = \frac{250}{1000} = 0.25\,mol/L
To find the mass needed, use the formula: Mass (g)=Concentration (mol/L)×Volume (L)×Molecular Weight (g/mol)\text{Mass (g)} = \text{Concentration (mol/L)} \times \text{Volume (L)} \times \text{Molecular Weight (g/mol)}
The molecular weight of potassium chloride (KCl) is 74.55 g/mol. Thus, the calculation is: 0.25 mol/L×0.015 L×74.55 g/mol=0.27956 g0.25\,mol/L \times 0.015\,L \times 74.55\,g/mol = 0.27956\,g
Convert this to milligrams: 0.27956 g×1000=279.56 mg0.27956\,g \times 1000 = 279.56\,mg
Hence, 279.56 mg of KCl is required.
2. Preparing 75 ml of an 800 µM KCl solution:
Similarly, first convert volume to liters: 75 ml=751000=0.075 L75\,ml = \frac{75}{1000} = 0.075\,L
Convert micromolar (µM) to molar (M): 800 µM=8001,000,000=0.0008 mol/L800\,µM = \frac{800}{1,000,000} = 0.0008\,mol/L
Using the same formula: 0.0008 mol/L×0.075 L×74.55 g/mol=0.004473 g0.0008\,mol/L \times 0.075\,L \times 74.55\,g/mol = 0.004473\,g
Convert this to milligrams: 0.004473 g×1000=4.47 mg0.004473\,g \times 1000 = 4.47\,mg
Thus, only 4.47 mg of KCl is required for the second solution.
The difference in mass required between these two solutions highlights the importance of precise calculations for accurate concentrations in laboratory settings.