# If it takes #"3 min"# for #40%# of sucrose to dissolve in a certain amount of water, how long does it take for all but #10%# of the sugar to dissolve?

##### 1 Answer

Not exactly a half-life problem... but it seems we'll have to assume that it works that way (even though it is more random than that).

Here are three ways to do it.

**CONCEPTUAL WAY**

A **half-life** is the amount of time needed to see the concentration drop to one half its starting amount. We could represent this as:

#[suc] = (1/2)^(n)[suc]_0# where

#n# is the number of "half-lives" (the number of times we've multiplied the starting concentration by#0.5# ).

Extending the idea of a half-life, the *"0.4-life"* can be represented as:

#bb([suc] = (0.4)^(n)[suc]_0)# where

#n# is the number of "0.4-lives" (the number of times we've multiplied the starting concentration by#0.4# ). Each "0.4-life" is#"3 min"# , as defined in the question.

If we then let

#=> 0.1cancel([suc]_0) = 0.4^n cancel([suc]_0)#

#=> 0.1 = 0.4^n#

#=> ln0.1 = ln 0.4^n#

Use the property that

#=> ln0.1 = nln0.4#

#=> n = (ln0.1)/(ln0.4) = 2.513#

That means about

**USING EXCEL?**

If you wanted to try doing this with excel, what I did was choose

Then, I used the equation above,

And the amount of time needed to reach

**USING THE INTEGRATED RATE LAW**

** Assuming** sugar solvation follows first-order half-life kinetics, another way is to use the

**first-order integrated rate law**:

#bb(ln[suc] = -kt + ln[suc]_0)#

Rearrange to get:

#ln(([suc])/([suc]_0)) = -kt#

In the question, it states that

#ln((0.4cancel([suc]_0))/(cancel([suc]_0))) = -k(3)#

#ln0.4 = -3k#

And the **rate constant** for this solvation process would be:

#=> k = -ln0.4/3 = "0.3054 min"^(-1)#

As a result, if we want all but

#ln((0.1cancel([suc]_0))/(cancel([suc]_0))) = -("0.3504 min"^(-1))t#

#=> ln 0.1 = -0.3504t#

#=> color(blue)(t) = -ln0.1/0.3054 "min"#

#=# #color(blue)("7.54 min")#

You can see we got the same thing as we got above.