Question

1- 0.08 moles of lithium fluoride in 40 mL of solution

2- 100 grams of NaOH in 4 liters of solution

3- what volume of a 2.0 M solution is needed to provide 0.5 mole of NaOH

4- what is the molarity of 42.5 grams of NH3 in 0.5 L of solution

Answer 1

1.) lithium fluoride LiF

40ml = 0.040L

M = $\frac{0.08}{0.040}$ = 2 M

2.) NaOH

MW_NaOH = 39.997g/mol

n = 100g ($\frac{1mol}{39.997g}$ ) = 2.50 mol

M = $\frac{2.50mol}{4L}$ = 0.625M

3.) NaOH

M = $\frac{0.5mole}{volume}$ , M = 2.0M

2 = $\frac{0.5}{volume}$

volume = $\frac{0.5}{2}$ = 0.25L

4.) NH₃

MW_NH3 = 17.031g/mol

n = 42.5 g ($\frac{1mol}{17.031g}$ ) = 2.50 mol

M = $\frac{2.5mol}{0.5L}$ = 5M