Considering that scientific notation has to move the decimal over so there is only one digit in the Tens place, that is the scientific notation.
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Answer:
The percentage by mass of benzene in the solution is approximately 0.2%
Explanation:
The given parameters are;
The mass of the benzene solute dissolved in the gasoline solvent, m₁ = 1.56 g
The total volume of the benzene gasoline solution made, V = 998.44 mL
The density of gasoline, ρ = 0.7489 g/mL
Mass, m = Density, ρ × Volume, V
∴ The mass of gasoline in the 998.44 mL, solution = 0.7489 g/mL × 998.44 mL = 747.731716 g
The total mass of the solution = The mass of the benzene in the solution + The mass of the gasoline in the solution
∴ The total mass of the solution = 747.731716 g + 1.5 g = 749.231716 g

The percentage by mass of benzene in the solution = (1.5 g/749.231716 g)×100 ≈ 0.2% by mass.
To determine the molar mass of the unknown gas, we use Graham's Law of Effusion where it relates the effusion rates of two gases with their molar masses. It is expressed as r1/r2 = √M2/M1. We calculate as follows:
Let 1 = argon gas 2 = unknown gas
r2 = 0.91r1r1/r2 = 1/0.91
1/0.91 = √M2/M1 = √M2/40M2 = 48.30 g/mol
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students