Question

How many grams of NaCl are needed to prepare 30 mL of a 0.87 M NaCl solution?

Answer 1

How many grams of NaCl are needed to prepare 30 mL of a 0.87 M NaCl solution?

This type of problem is solved using stoichiometric ratios and unit conversion. The first thing you should find is how many moles are dissolved in the 30 ml of solution you want to prepare. To do this, you must make a ratio with this volume and the concentration of said solution:

The concentration of the solution is molar, that means that for every liter of solution, 0.87 moles of the solute are available. Knowing this, the following relationship is established: If in 1000 mL of solution (which is the equivalent of 1 liter) you have 0.87 moles of NaCl, how many moles of NaCl will you have in 30 mL?

1000 mL sol ---------> 0.87 moles of NaCl
30 mL sol ---------> X moles of NaCl

moles NaCl = (30 * 0.87) / 1000 = 0.0261 moles NaCl

Note that the volume units are canceled due to division, leaving the moles of the solute. The result tells you that 0.0261 moles of NaCl are needed to prepare 30 ml of solution. Now what you should do is convert those moles to grams using the molecular mass of NaCl, which is 58.4 g/mol

1 mol NaCl -------> 58.4 g NaCl
0.0261 mol NaCl -------> X

g NaCl = (0.0261 * 58.4) / 1 = 1.52424 g of NaCl

So, you need 1.52424 g of NaCl to prepare 30 mL of a 0.87 M NaCl solution.