Question

A solution of rubbing alcohol is 79.3 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 79.0  mL sample of the rubbing alcohol solution?

How many liters of a 3.90  M K2SO4 solution are needed to provide 90.8  g of K2SO4 (molar mass 174.01 g/mol)? Recall that M is equivalent to mol/L.

Answer 1

1)

(%v/v) = volume of isopropanol * 100 / volume of sample

79.3 = volume of isopropanol*100/79.0

volume of isopropanol = 62.6 mL

Answer: 62.6 mL

2)

Molar mass of K2SO4 = 174.01 g/mol

mass of K2SO4 = 90.8 g

we have below equation to be used:

number of mol of K2SO4,

n = mass of K2SO4/molar mass of K2SO4

=(90.8 g)/(174.01 g/mol)

= 0.522 mol

we have below equation to be used:

M = number of mol / volume in L

3.9 = 0.522/ volume in L

volume = 0.134 L

volume = 134 mL

Answer: 134 mL