How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?

Options

A. 60g
B. 6g
C. 4g
D. 40g

Answer

B

Explanation

`1500 cm^(3)` of `0.1N HCI -= 1500 cm^(3)` of `0.1N NaOH`
`-= 1500 cm^(3)` of `0.1M NaOH`
`=((1500)/(1000)L)xx (0.1 mol L^(-1)) = 0.15` mol
`= 0.15` mol `xx 40 g mol^(-1) = 6g`