A solvent contains 76% as X and left out 24%as Y (of concentration 50%)
if we add Y of (75% conc) .in what proportion we have to add X and Y keeping conc of X constant
2006-07-06 22:10:15 · 6 answers · asked by rakesh_506 1 in Science & Mathematics ➔ Chemistry
Ummmm.....so x is 1/8 of the solution (25% of 50%) and y is 3/8 of the solution (75% of 50%).
That is a 3:1 ratio for (y):(x) [or 1:3 ratio of (x):(y)]
Well, if (y) = 3(x), [or (x) = 1/3(y)], then whatever percentage of (y) you add, (x) must be 1/3 of that. OR whatever percentage (x) you add, (y) must be 3 times that.
2006-07-06 22:28:30 · answer #1 · answered by joshua2778 3 · 0⤊ 0⤋
I undrerstand that this is a dilution problem, or alliqout. But your manner of asking is kinda vague..."in what proportion we have to add X and Y" - I thought you want to add the Y of the 75% to the X and Y of the 50% solution, not the other way around? It matters which is to be added to which, especially in mixing emulsions. And another thing, "keeping conc of X constant" - X of what, the 50% solution? You gotta ask clearly so we could clarify whatever it is that you wanna ask :)
Without using the formula for dilution and by virtue of philosophy, the question "in what proportion we have to add X and Y keeping conc of X constant" would be best answered with "the same proportion, 76% of X in the 50% solution" since you only mentioned one value of X in your problem, so I assumed your talking about the same X, the one and only X in your problem, and you were just trying to mislead us. Did I get it right? :p
2006-07-07 06:52:02 · answer #2 · answered by gameplan_xtreme 4 · 0⤊ 0⤋
Kentucky.
2006-07-07 06:30:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1/4 of hahaha + 3/4 of lol
2006-07-07 05:15:12 · answer #4 · answered by Axel ∇ 5 · 0⤊ 0⤋
18% of Y shoul be added,I think.
2006-07-07 12:13:56 · answer #5 · answered by Naema B 1 · 0⤊ 0⤋
omg i dont know what ur on about~!
2006-07-07 05:15:50 · answer #6 · answered by Anonymous · 0⤊ 0⤋