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How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?A. 60gB. 6gC. 4gD. 40g |
Answer» Correct Answer - B `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` |
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