A solution of the solute hexane and the solvent acetone that is 28.1 g hexane per 100. g solution is available. Calculate the mass (g) of the solution that must be taken to obtain 0.646 kg of CH3(CH2)4CH3.
Molar Mass (g/mol)
CH3(CH2)4CH3 100.21
(CH3)2CO 58.05
Density (g/mL):
CH3(CH2)4CH3 0.6838
(CH3)2CO 0.7899
Name/Formula:
hexane
CH3(CH2)4CH3
acetone
(CH3)2CO
You have a solution that is 28.1 g solute/100 g solution. You want 646 g of the solute. Wouldn't that just be
646 x (100/28/1) = ?? OR
by ratio/proportion
(28.1/100) = (646/x)
x = 646 x 100/28.1 = ??
You can check it to make sure it is ok. If I do the math I get close to 2299 g solution needed.
So 646g solute/2299 g soln = 0.281/g or 28.1/100g
Check my thinking. Check my work.
To calculate the mass of the solution needed to obtain 0.646 kg of CH3(CH2)4CH3, you can follow these steps:
Step 1: Convert the given mass of CH3(CH2)4CH3 from kg to grams.
0.646 kg = 646 grams
Step 2: Calculate the number of moles of CH3(CH2)4CH3.
Number of moles = mass / molar mass
Number of moles = 646 g / 100.21 g/mol
Number of moles ≈ 6.4522 moles
Step 3: Use the mole ratio from the balanced equation to find the number of moles of hexane needed.
The balanced equation is:
CH3(CH2)4CH3 + (CH3)2CO ⇒ Solution
From the equation, we can see that the mole ratio between CH3(CH2)4CH3 and hexane is 1:1.
Step 4: Calculate the mass of hexane needed.
Mass of hexane = moles of hexane × molar mass of hexane
Molar mass of hexane = 100.21 g/mol
Mass of hexane = 6.4522 moles × 100.21 g/mol
Mass of hexane ≈ 646.4 g
Step 5: Calculate the mass of the solution needed.
The given solution is 28.1 g of hexane per 100 g of solution. This means that the proportion of hexane in the solution is 28.1/100 = 0.281.
Let x be the mass of the solution needed.
Mass of hexane in the solution = x × 0.281 g
Since we know that the mass of hexane needed is approximately 646.4 g, we can set up the equation:
x × 0.281 g = 646.4 g
Finally, solve for x:
x ≈ 646.4 g / 0.281 ≈ 2302.9 g
Therefore, the mass of the solution that must be taken to obtain 0.646 kg of CH3(CH2)4CH3 is approximately 2302.9 grams.