- #1
Lucy345
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Homework Statement
Known: [/B]
Mass of compound A: 15mg (weighed in excess), MM=200g/mol
Mass of compound B: 6mg, MM=700g/mol
Concentration of each compound listed above(Concentration of solution of compound A or solution of Compound B): 20mg/L
Total concentration of the solution: 20mg/mL
Molar percent(percent of total number of moles in solution) of compound A: 20%
Problem:
I will list everything that is given to me by the problem as (Given:)
So I measured a mass of compound A, let's say 15mg. And I measure a mass of compound B, let's say 6mg. (Given: concentration of each compound is 20mg/mL). Compound A is measured in excess. So using the mass, and the given concentration, I calculate the volume required for each compound(using the given concentration of 20mg/mL) and mix each compound into their required volumes to create 2 solutions. However, (Given: 10% molar percent[that is to say 10% of the total number of moles in the final solution]) of Compound A, what is the mass and volume of Compound A that I need to add to Compound B in order to get that mole percent? However, I cannot change the total concentration of the solution(Given: 20mg/mL)
2. Homework Equations
C=nv etc simple chem equations
The Attempt at a Solution
Moles A: 15mg/1000/200g/mol=7.5*10^(-5)
Moles B: 6mg/1000/700g/mol=8.57*10^(-7)
Total Moles in sol'n*(1-0.2)=Moles Compound B
Total Moles in sol'n=1.07*10^(-6) moles
Moles of Compound A needed=1.07*10^(-6) moles*0.2=2.14*10^(-7)
Mass of Compound A=2.14*10^(-7)*200g/mol*1000=0.043mg
Volume of Compound A to add=0.043mg/(20mg/mL)=0.00214mL
^This answer is wrong however.