In this particular case weighing would work with limited accuracy (I am going to explain the limitations later in the answer).
Since the only compound leaving the reaction is the unreacted $\ce{CO2}$, it is valid to use weighing. Everything else is either sticking around or is already included in the initial weighing. So by weighing everything prior to reaction and after "all unreacted $\ce{CO2}$ is vaporized", you can calculate the amount of "reacted $\ce{CO2}$".
(weight of reacted $\ce{CO2}$) = (Weight of initial $\ce{CO2}$) - (weight of unreacted $\ce{CO2}$)
(Weight of initial $\ce{CO2}$) = (50 g) - (condensed water from atmosphere [estimate])
(Weight of unreacted $\ce{CO2}$) = (deficit in pre vs post reaction weighing)
pre reaction weight: 50 g $\ce{CO2}$ + 25 g $\ce{KOH}$ +100 g water = 175 g
post reaction weight: weight of the whole solution after all unreacted $\ce{CO2}$ is escaped from the solution (make sure the whole solution is at room temperature and make sure there is no bubbling occurring).
I understand there are some practical limitations, mainly :
1- There might be unreacted/frozen $\ce{CO2}$ still hanging around, Try to shake well. Also when weighing post reaction, make sure your solution is slightly warm to assist unreacted $\ce{CO2}$ to evaporate fully from the reaction.
2-Since you have dry ice, some water from atmosphere will condense into your reaction, this will lead to underestimation of the "leaving compound". The way to overcome, is to keep your reaction vessel as small as possible and control its inlet by a cold trap to make sure no water from atmosphere gets in. Also account for amount of water usually present in 50 g dry ice, by examining the water content on a similar sample beforehand.
pH:
Your initial pH is clearly not right. concentration of your $\ce{OH-}$ is way higher than just 0.1 M. Let's assume that it is due to high pH value being out of the calibration range of the pH-meter and let's assume that your pH=10 after addition of $\ce{CO2}$ is correct.
One will be able to calculate the amount of unreacted $\ce{CO2}$ from reaction balance. You are starting with 25 g (0.446 moles) of $\ce{KOH}$ and 50 g (1.14 moles) of $\ce{CO2}$. For every $\ce{CO2}$ reacting, 2 $\ce{KOH}$ are used up. So IF all(or most) of $\ce{KOH}$ is used up, you should end up with 0.446/2~ 0.22 moles of $\ce{K2CO3}$, which means the reacted CO2 is about 0.22 moles(~9.7 g), and unreacted is 0.92 moles(~40.3 g).
Why do I make the assumption that all the $\ce{KOH}$ is reacted, notice your final pH is 10 , meaning that your [OH-] concentration is about 0.0001 M, your starting concentration was 4.4 M.
What stops the reaction from using all $\ce{KOH}$ to be used up and dropping the pH to neutral zone? You are using excess $\ce{CO2}$, so we expect all the $\ce{KOH}$ to be used up. This has to do with part of $\ce{CO2}$ escaping the reaction as gas, also since the temperature is freezing the water (by dry ice) you may not have complete agitation which hinders the reaction procession to some extent.
I still think that you should take the weighing method seriously. Using weighing methods will provide confirmation for your calculations.
