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The mass of 60% HCl by mass required for the neutralisation of 10 L of `0*1` M NaOH isA. `60*8 g`B. `21*9 g`C. `100 g`D. `219 g` |
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Answer» Correct Answer - A `underset(36.5g)(HCl)+underset(40g)(NaOH)toNaCl+H_(2)O` Amount of NaOH to be neutralised `=0*1xx10xx40=40g` `:.` HCl required `= 36*5 g` 60% HCl required `= (36*5)/(60)xx100=60*8g` |
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