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 $ .



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