Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | Chemistry
One Answer Below
Anyone can contribute an answer, even non-tutors.
Please help me as I do not understand what part a is saying. (It may be of a difficulty higher than sec 3 cuz I’m in IP)
Date Posted: 3 years ago
Views: 742
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Hope this helps :)
Date Posted:
3 years ago
Part a is asking for the number of moles of Na2CO3 in 250cm^3 of FA1 solution, although they mention in the earlier part of the question that they carry out titration with only 25cm^3 of Na2CO3 (NOT 250cm^3). So how do you find the number of moles of Na2CO3 in 250cm^3 when they only use 25cm^3 in the reaction?
First, you need to find the number of moles of Na2CO3 in 25cm^3 of FA1 in order to find out how many moles of Na2CO3 is present in 250cm^3. You can do this by comparing the ratio of the volumes of Na2CO3 (250:25 = 10:1) which explains why in the final step, you need to multiply by 10 to the no. of moles of Na2CO3 in 25cm^3 of FA1. You can use this ratio method because mole ratio is proportional to volume ratio. You can ignore the no. of moles of NaCl completely because NaCl doesn't take part in the reaction.
First, you need to find the number of moles of Na2CO3 in 25cm^3 of FA1 in order to find out how many moles of Na2CO3 is present in 250cm^3. You can do this by comparing the ratio of the volumes of Na2CO3 (250:25 = 10:1) which explains why in the final step, you need to multiply by 10 to the no. of moles of Na2CO3 in 25cm^3 of FA1. You can use this ratio method because mole ratio is proportional to volume ratio. You can ignore the no. of moles of NaCl completely because NaCl doesn't take part in the reaction.
360 Tutoring Program