Question

4) A 34.0 % (m/m) sulfuric acid (H2SO4) solution has a density of 1.25 g/mb. What is the molarity of this solution?

Answer 1

Answer:

Step 1: Explanation

Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%

i.e   % m/m = ( Mass  of solute / Mass of solution ) x 100

Step 2: Calculate of moles of sulfuric acid

Given,

34% (m/m) which means it contains 34 g of sulfuric acid in 100 g of solution

hence,

mass of sulfuric acid = 34 g

mass of solution = 100 g

we know, molar mass of sulfuric acid (H2SO4 ) = 98.079 g/mol

Moles of sufuric acid = mass given / molar mass = 34 g / 98.079 g/mol = 0.34666 mol

Step 3: Calculation of Volume of solution

Given,

density of solution = 1.25 g/mL

Mass = 100 g

we know,

density = mass / volume

=> volume = mass / density

=> volume = 100 g / 1.25 g/mL = 80 mL

Hence, volume = 80 mL = 0.08 L

[ because 1 L=1000 mL so, 80 mL = ( 80 mL  × ( 1 L / 1000 mL) = 0.08 L ]

Step 4: Calculation of Molarity of solution

Molarity is defined to be the number of moles of solute divided by volume of solution in liters

Molarity(M)= number of moles of solute / volume of solution ( in L )

We got,

Moles of sulfuric acid = 0.34666 mol

Volume of solution = 0.08 L

on substituting the value

molarity = moles of solute / volume of solution in L

=> molarity = 0.34666 mol /0.08 L = 4.33 M [ note : mol /L = M (unit of molarity ) ]

Hence, the molarity is 4.33 M