Question

 

1. How much Sodium chloride (NaCl), in grams, would you need to prepare 100 mL of a 0.2 M NaCl solution, giving the molecular weight for NaCl is 58.4 g/mole?

2. How much glucose, in grams, would you need to prepare 25 mL of a 24% glucose solution?

3. How much stock solution would you need to make 50 mL of a 86 mM Potassium chloride (KCl) solution, given the stock solution is 1 M?

Answer 1

1. Molarity of solution M = n x 1000 / V (mL)

where, n = moles of solute NaCl, V = volume of solution (mL), given V = 100 mL and M = 0.2M

Now, moles of NaCl, n = M X V / 1000

n = 0.2 x 100 / 1000

n = 0.02 mole

Hence, mass of NaCl = n x molecular weight

mass of NaCl = 0.02 x 58.4

mass of NaCl = 1.168g

2. % strenght of solution (w/v) = mass of solute (g) x 100 / volume of solution

given, volume = 25 mL, % strength = 24%

So, mass of solute (glucose) = % strength x volume of solution / 100

mass of solute (glucose) = 24 x 25 / 100

mass of solute (glucose) = 6.0g

3. Use the relation C1V1 = C2V2

given, C1 = 1M, V1 = ?, C2 = 86mM = 0.086M and V2 = 50 mL

So, V1 = C2V2 / C1

V1 = 0.086 x 50 / 1

V1 = 4.3 mL