NEET Exam  >  NEET Questions  >  3.5g of a mixture of NaOH and KOH were dissol... Start Learning for Free
3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is?
Most Upvoted Answer
3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL...
Calculating the Percentage of KOH in Mixture


To calculate the percentage of KOH in the mixture, we need to follow the below steps:


Step 1: Calculate the number of moles of HCl used


We have used 17 mL of (N/2) HCl solution for neutralizing 25 mL of the mixture. The concentration of (N/2) HCl solution is:


(N/2) HCl = (1/2) x 36.5 g/mol = 18.25 g/mol


Now, the number of moles of HCl used can be calculated as:


Number of moles of HCl = (Concentration x Volume) / 1000


Number of moles of HCl = (18.25 x 17) / 1000 = 0.31025 moles


Step 2: Calculate the number of moles of NaOH and KOH in the mixture


Let us assume that the percentage of KOH in the mixture is x%. So, the percentage of NaOH in the mixture will be (100 - x)%.


Now, the number of moles of NaOH and KOH in the mixture can be calculated as:


Number of moles of NaOH = (Weight of NaOH / Molar mass of NaOH)


Number of moles of KOH = (Weight of KOH / Molar mass of KOH)


Weight of NaOH = x/100 x 3.5 g


Weight of KOH = (100 - x)/100 x 3.5 g


Molar mass of NaOH = 40 g/mol


Molar mass of KOH = 56 g/mol


Number of moles of NaOH = (x/100 x 3.5) / 40 = 0.00875 x


Number of moles of KOH = [(100 - x)/100 x 3.5] / 56 = 0.00625 (100 - x)


Step 3: Calculate the number of moles of KOH neutralized by HCl


The number of moles of KOH neutralized by HCl can be calculated by subtracting the number of moles of NaOH from the total number of moles of HCl used:


Number of moles of KOH neutralized = Number of moles of HCl used - Number of moles of NaOH


Number of moles of KOH neutralized = 0.31025 - 0.00875 x


Step 4: Calculate the percentage of KOH in the mixture


Now, we can use the number of moles of KOH neutralized to calculate the percentage of KOH in the mixture:


Number of moles of KOH neutralized = 0.00625
Community Answer
3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL...
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
Explore Courses for NEET exam

Top Courses for NEET

3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is?
Question Description
3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is?.
Solutions for 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? in English & in Hindi are available as part of our courses for NEET. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? defined & explained in the simplest way possible. Besides giving the explanation of 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is?, a detailed solution for 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? has been provided alongside types of 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? theory, EduRev gives you an ample number of questions to practice 3.5g of a mixture of NaOH and KOH were dissolved and made up to 250 mL. 25 mL of this solution were completely neutralised by 17 mL of (N/2) HCl solution. Then, the percentage of KOH in mixture is? tests, examples and also practice NEET tests.
Explore Courses for NEET exam

Top Courses for NEET

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev