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How many grams of NaCl are required to make 250.0 mL of a 3.000 M solution?

A. 58.48 g
B. 175.3 g
C. 14.68 g
D. 43.83 g
E. 32.36 g


Sagot :

To determine the number of grams of NaCl required to prepare a 250.0 mL solution with a molarity of 3.000 M, follow these steps:

1. Understand the given variables:
- Molarity ([tex]\( M \)[/tex]) is 3.000 mol/L.
- Volume ([tex]\( V \)[/tex]) is 250.0 mL.
- The molar mass of NaCl is 58.44 g/mol.

2. Convert the volume from milliliters to liters:
[tex]\[ \text{Volume in liters} = \frac{\text{Volume in mL}}{1000} = \frac{250.0}{1000} = 0.250 \text{ L} \][/tex]

3. Calculate the number of moles of NaCl required:
[tex]\[ \text{Moles of NaCl} = \text{Molarity} \times \text{Volume in liters} = 3.000 \text{ mol/L} \times 0.250 \text{ L} = 0.750 \text{ mol} \][/tex]

4. Convert the moles of NaCl to grams using the molar mass:
[tex]\[ \text{Mass of NaCl} = \text{Moles of NaCl} \times \text{Molar mass of NaCl} = 0.750 \text{ mol} \times 58.44 \text{ g/mol} = 43.83 \text{ g} \][/tex]

Thus, to make 250.0 mL of a 3.000 M NaCl solution, you need 43.83 grams of NaCl.