What are the calculations to find the number of moles needed to prepare 10mL of a 0.2M 2-butenedioic acid (C4H4O4)?
Number of moles required to prepare 10mL of a 0.2M 2-butenedioic acid `(C_4H_4O_4) = ((1 mol)/(1L))/(1M)`` ``xx(1L)/(1000mL)xx10 mLxx0.2M`
` =0.002 mol =2.0xx10^(-3) mol`
Turn the 10ml in L since the equation for Molarity is M=mol/L and we need to find mols okay-
10mLx(1mL/1000L) = 0.01L Now plug into the equation
0.20 M C4H404 = x mol/ 0.01L Now multiply 0.01L on both sides
0.002 mole C4H404= x mol Since we have mols now we have to find the amount of grams need so calculate the molar mass of the entire compound.
x mol =grams/molar mass Plug in!
0.0002 mole C4H404= x grams/ 116g Multiply on both sides
x grams =0.232g and 0.232grams is the amount need to prepare C4H404
10ml = 0.01L (match the units)
M = mol / L (this is basically what M means)
= 0.2mol/L C4H4O4 = mol C4H4O4 / 0.1L C4H4O4 (plug in the known information into the equation)
= mol C4H4O4 = 0.2mol/L C4H4O4 * 0.1L C4H4O4 (multiply the mol on the left and M to the right)
= 0.002mol C4H4O4 (final answer)
= 2.0 * 10^-3
Hope this helped!
Molarity (M) is concentration of a solution in terms of moles of solute/ liters of solution.
0.2 M is the same as `(0.2 mol C _4H_4O_4)/(1 L of solution)`
The new solution will be ` (X mol C_4H_4O_4)/ (0.010 L)`
Note 1000 mL = 1L then `10 mL *( (1 L)/(1000 mL))= 0.010 L `
Applying ratio and proportion:
`((0.2 mol C_4H_4O_4)/(1 L)) = ((X mol C_4H_4O_4)/(0.010 L))`
Cross-multiply 0.010 L to isolate X mol` C_4H_4O_4` :
X mol `C_4H_4O_4` =`((0.2 mol C_4H_4O_4)/(1 L))* (0.010 L)`
X mol `C_4H_4O_4` = 0.002 mol `C_4H_4O_4`
or `2x10^(-3) mol C_4H_4O_4`
In order to do this problem it will be helpful to remember that:
`M = (mol)/(L)` therefore is we are looking for moles it is: `Mol = MxxL`
We know the L = 10ml (.01L) and the M = .2 and we just set it up:
Moles = `(.2)xx(.01)`
Moles = .0002 Mol or `2 xx 10^-3 ` mol