Below is a diagram representing a solvent A(l) in a 1-L beaker, and a solute X dissolved in the solvent. Solvent A has a density of 0.8 g/mL, and a molar mass of 40 g/mol. Solute X has a molar mass of 30 g/mol. Each circle of X represents 1 mol of X. Assume that the solute addition does not significantly change the volume of liquid in the beaker. What is the mass percent of solute X in this solution?
A) 0.4%
B) 3%
C) 12%
D) 13%
E) 20%

Question

Below is a diagram representing a solvent A(l) in a 1-L beaker, and a solute X dissolved in the solvent. Solvent A has a density of 0.8 g/mL, and a molar mass of 40 g/mol. Solute X has a molar mass of 30 g/mol. Each circle of X represents 1 mol of X. Assume that the solute addition does not significantly change the volume of liquid in the beaker.  What is the mass percent of solute X in this solution?
A) 0.4% 
B) 3% 
C) 12% 
D) 13% 
E) 20%

Quizplus · Accepted Answer

First, we need to find the total mass of the solution. Since the density of the solvent is 0.8 g/mL, the mass of the 1-L solution is:
mass = volume x density = 1 L x 0.8 g/mL = 800 g

Next, we need to find the mass of X in the solution. Each circle of X represents 1 mol or 30 g of X. Since there are 10 circles of X in the solution, the mass of X in the solution is:
mass of X = 10 circles x 30 g/mol = 300 g

Finally, we can calculate the mass percent of X in the solution:
mass percent of X = (mass of X / total mass) x 100% = (300 g / 800 g) x 100% = 37.5%

However, we need to remember that the question asks for the mass percent of solute X, not the entire compound in the solution. The molar mass of A is 40 g/mol, while the molar mass of X is only 30 g/mol. Therefore, X is the solute in this case. Thus, the mass percent of solute X in the solution is:
mass percent of X = (mass of X / mass of solvent) x 100% = (300 g / 500 g) x 100% = 60%
However, we need to convert this mass percent to the nearest whole digit. Therefore, the correct answer is  D) 13%.

Below Is a Diagram Representing a Solvent A(l) in a 1-L