A second-year chemistry student is asked to make up a 250.0 mL sample of 5.87 M aqueous sodium chloride, NaCl solution.Molar mass of NaCl = 58.44 g mol-1 (a) Calculate the mass (in grams) of solid sodium chloride that the student would need to weigh out. Provide your numerical answers and units. (b) The student is then asked to make up 100.0 mL of sodium chloride solution using 15.00 mL of the 5.87 M NaCl solution made up in part (a). The diluted solution is then mixed completely. Determine the concentration of the diluted sodium chloride solution.

(a) To calculate the mass of solid sodium chloride (NaCl) needed to prepare a 250.0 mL sample of 5.87 M NaCl solution, we can use the formula:

$\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$

First, convert the volume from mL to L:

$250.0 \, \text{mL} = 0.2500 \, \text{L}$

Next, use the molarity to find the moles of NaCl:

$\text{Moles of NaCl} = \text{Molarity} \times \text{Volume in liters}$

$\text{Moles of NaCl} = 5.87 \, \text{M} \times 0.2500 \, \text{L}$

$\text{Moles of NaCl} = 1.4675 \, \text{moles}$

Now, convert moles to grams using the molar mass of NaCl (58.44 g/mol):

$\text{Mass of NaCl} = \text{Moles of NaCl} \times \text{Molar mass of NaCl}$

$\text{Mass of NaCl} = 1.4675 \, \text{moles} \times 58.44 \, \text{g/mol}$

$\text{Mass of NaCl} = 85.75 \, \text{g}$

So, the student would need to weigh out 85.75 grams of solid sodium chloride.

(b) To determine the concentration of the diluted sodium chloride solution, we can use the dilution formula:

$C_1 V_1 = C_2 V_2$

Where:

• $C_1$ is the initial concentration (5.87 M)
• $V_1$ is the initial volume (15.00 mL)
• $C_2$ is the final concentration
• $V_2$ is the final volume (100.0 mL)

First, convert the volumes to liters:

$V_1 = 15.00 \, \text{mL} = 0.01500 \, \text{L}$

$V_2 = 100.0 \, \text{mL} = 0.1000 \, \text{L}$

Now, use the dilution formula to find $C_2$:

$(5.87 \, \text{M}) \times (0.01500 \, \text{L}) = C_2 \times (0.1000 \, \text{L})$

$0.08805 \, \text{mol} = C_2 \times 0.1000 \, \text{L}$

$C_2 = \frac{0.08805 \, \text{mol}}{0.1000 \, \text{L}}$

$C_2 = 0.8805 \, \text{M}$

So, the concentration of the diluted sodium chloride solution is 0.8805 M.