To determine the molality of sulphuric acid in the solution, we'll follow these steps:
1. First, we need to use the given molarity (M) and density (ρ) to find the mass of the solution.
Given:
Molarity
(M)=4.5MDensity
(ρ)=1.28g∕ml 2. The molarity
(M) is defined as the number of moles of solute (sulphuric acid,
H2SO4 ) per liter of solution. We need to find out how many grams of
H2SO4 are present in 1 liter of solution:
Number of moles of
H2SO4 in 1 liter of solution
=4.5molMolar mass of
H2SO4=2×1+32+4×16=98g∕molTherefore, mass of
H2SO4 in 1 liter of solution
=4.5×98=441g 3. Next, we will determine the total mass of the solution:
Since density
(ρ) is the mass per unit volume, the mass of 1 liter
(1000ml) of solution is:
Mass of solution
=1.28g∕ml×1000ml=1280g 4. The mass of the solvent (water) in the solution can be found by subtracting the mass of the solute (sulphuric acid) from the total mass of the solution:
Mass of water
=1280g−441g=839g 5. Now, we use the definition of molality
(m), which is the number of moles of solute per kilogram of solvent (water):
Mass of water in kilograms
==0.839kgTherefore, the molality
(m) is:
Therefore, the molality ( m ) is:
m==5.364mol∕kgThus, the molality of the acid is
5.364mol∕kg.
So, the correct answer is: Option C - 5.364