The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol–1) by mass will b ?

To determine the density of a 3.60 M sulfuric acid solution that is 29% H2SO4 by mass, we need to break down the problem into manageable steps. We’ll use the definitions of molarity, mass percent, and the relationship between mass, volume, and density. Let’s dive into the calculations.

Understanding Molarity and Mass Percent

Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, we have a 3.60 M solution of sulfuric acid (H2SO4). This means there are 3.60 moles of H2SO4 in every liter of the solution.

Mass percent tells us how much of the solution's mass is made up of the solute. A 29% mass percent means that in 100 grams of the solution, 29 grams are H2SO4.

Calculating the Mass of H2SO4 in the Solution

First, we calculate the mass of H2SO4 in one liter of the solution:

• Since the molar mass of H2SO4 is 98 g/mol, the mass of H2SO4 in 3.60 moles is:
• Mass of H2SO4 = 3.60 moles × 98 g/mol = 352.8 grams

Finding the Total Mass of the Solution

Next, we need to find the total mass of the solution. Since we know that 29% of the solution's mass is H2SO4, we can set up the following equation:

Let \( x \) be the total mass of the solution. Then:

• 0.29x = 352.8 grams
• Solving for \( x \):
• x = 352.8 grams / 0.29 ≈ 1215.86 grams

Calculating the Density of the Solution

Density is defined as mass per unit volume. We know the mass of the solution (approximately 1215.86 grams) and the volume of the solution (1 liter or 1000 mL). Therefore, we can calculate the density:

Density = Mass / Volume

• Density = 1215.86 grams / 1000 mL = 1.21586 g/mL

Final Result

Thus, the density of the 3.60 M sulfuric acid solution that is 29% H2SO4 by mass is approximately 1.22 g/mL.

This calculation illustrates how we can use molarity and mass percent to find the density of a solution. By understanding the relationships between these concepts, we can solve various problems in chemistry effectively.