Solutions Question 361
Question: How much of NaOH is required to neutralise 1500 $ cm^{3} $ of 0.1 N HCl (At. wt. of Na =23) [KCET 2001]
Options:
A) 4 g
B) 6 g
C) 40 g
D) 60 g
Show Answer
Answer:
Correct Answer: B
Solution:
1500 $ cm^{3} $ of 0.1 N HCl have number of gm equivalence $ =\frac{N_1\times V_1}{1000}=\frac{1500\times 0.1}{1000}=0.15 $
$ \therefore 0.15,gm. $ equivalent of NaOH $ =0.15\times 40=6,gm. $
BETA
