Answer:- 3.84 grams
Solution:- Volume of the sample is 44.8 mL and the density is 1.03 gram per mL.
From the density and volume we calculate the mass as:
mass = volume*density
= 46.1 g
From given info, potassium bromide solution is 8.34% potassium bromide by mass. It means if we have 100 grams of the solution then 8.34 grams of potassium bromide is present in. We need to calculate how many grams of potassium bromide are present in 46.1 grams of the solution.
The calculations could easily be done using dimensional analysis as:
= 3.84 g KBr
Hence, 3.84 grams of KBr are present in 44.8 mL of the solution.
<span>37.9968064 ± 0.0000010 g/mol</span>
Answer:
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Answer:
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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.
