JEE Main 12 Jan 2019 Morning Question 27
Question: 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 10 g
B) 80 g
C) 40 g
D) 20 g
E) None of these
Show Answer
Answer:
Correct Answer: E
Solution:
-
$ N _1V _1=N _2V _2 $ $ N _1 $ (oxalic acid) $ =2\times 0.5N $ $ 1\times 50=N _2\times 25\Rightarrow N _2=2N $
Normality $ =\frac{wt.}{eq.wt.}\times \frac{1000}{V(mL)} $
$ \Rightarrow $ $ 2=\frac{wt.}{40}\times \frac{1000}{50} $ weight = 4g
