How would you make up 100 mL of a 0.10 M NaCl solution?

Question

anonymous · Answer

Molarity = (moles of solute)/(Liters of solution)

$$M = \frac{ mol }{ L }$$
$$\frac{ 100mL }{ 1 }*\frac{ 1 L }{ 1000 mL }=0.1 L$$
$$0.1 M NaCl = \frac{moles NaCl }{ 0.1 Lsolution }$$
$$(0.1 M NaCl)(0.1 L solution) = moles NaCl$$
                                              = 0.01 moles NaCl
$$\frac{ 0.01 mol NaCl }{ 1 }*\frac{ 58 g NaCl }{ 1 mol NaCl }=0.58g NaCl$$
Add 0.58g solid NaCl, to 100 mL of H2O to make a 0.1 M NaCl solution.

anonymous · Answer

Thank you so much for being so clear!

anonymous · Answer

Ba(NO3)2 Molarity = moles of solute/Liters of solution ( 100 ml = 0.100 liters ) 0.10 M Ba(NO3)2 = moles Ba(NO3)2/0.100 liters = 0.01 moles Ba(NO3)2 (261.32 grams/1 mole Ba(NO3)2) = 2.6 grams of Ba(NO3)2 needs to be put into that 100 milliliters of solution.