The question is as follows:

went wrong, I'd really appreciate it!

Whoops, somehow I managed to mess that up. Here's the rest of my question

The question is as follows:
Find the initial concentration of a sucrose solution that has a current concentration of 3.45 molar after 2 hours and 30 minutes at 25°C. For this reaction at 25°C, k = 4.10 × 10^-2.

A. 3.82 molar
B. 3.82 molar
C. 3.11 molar
D. 3.79 molar

(Alright, third attempt here. My replies keep being eaten)

I inserted my givens into the equation, ' Natural-log At/Ao = negative constant multiplied by the time passed '. However, my answer isn't close to the choices. Any help would be appreciated!

What are the units on k = 4.10E-2 what?

If units are not available is this a first order reaction?
You have the same two answers for A and B? Is that right? If you will answer those question AND show your work I expect I can find your error. Assuming first order reaction I obtained one of the answers.

Sorry for the late reply!

Units for the constant are grams per liter (molar). Both answers A and B were exactly the same, so I assume there must have been a printing error, or something or the sort.

As for my work, I inserted the info given into the equation and solved from there:

In(At/Ao) = -k(t)
In(3.45/Ao) = -(4.1 X 10^-2)(150)
1.238 - In(Ao) = -6.15
-In(Ao) = -4.912
Ao = 1.5917

You didn't list the units of k but from your work I'm guessing that is in hours and you used minutes. When I substituted 2.5 hours in for time I obtained 3.82 M. That answer makes no sense when both A and B are listed as 3.82 M but that could be a printing error. Check to see that k has units of hours^-1 and if so then t must be substituted as 2.5 hours and not 150 minutes.

Ah! That would explain it. Silly oversight on my part, sorry about that.

K is defiantly in units of hours^-1, so that must be the correct choice. I'll check in with my teacher specifically tomorrow to see why the question ended up that way.

Thanks a ton for the help, I'll be sure to give more details in my question next time to check in for something like this!

I'm sorry, but I cannot answer your question without additional context. Could you please provide more information or rephrase your question?