Some Basic Concepts Of Chemistry Question 212
Question: How much of NaOH is required to neutralise 1500 $ cm^{3} $ of 0.1 $ N,HCl $ (Na = 23) [KCET 2001]
Options:
A) 40 g
B) 4 g
C) 6 g
D) 60 g
Show Answer
Answer:
Correct Answer: C
Solution:
$ N=\frac{\text{W(}gm)\times 1000}{V\times Eq\text{.wt}\text{.}} $
1500ml of 0.1N HCl = 150ml (N) $ 1=\frac{W(gm)\times 1000}{150\times 40} $ , W (gm) $ =\frac{150\times 40}{1000}=6gm $ .
