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A quantity of ammonium chloride was heated with 100 mL of 0.8 N NaOH solution till the reaction was complete. The excess of NaOH was neutralised with 12.5 mL of `0.75N H_(2)SO_(4)`. Calculate the quantity of ammonium chloride. |
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Answer» 12.5 mL of `0.75 N H_(2)SO_(4)-=12.5 mL " of" 0.75 N NaOH` 12.5 mL of 0.75 N NaOH `-= 11.72 mL " of" 0.8 N NaOH` NaOH solution used by `NH_(4)Cl` `=(100-11.72)mL ` of 0.8N NaOH =88.28 mL of 0.8 N NaOH `-=88.28 mL " of " 0.8 N NH_(4)Cl` Mass of `NH_(4)Cl` present in 88.28 mL of `0.8N NH_(4)Cl` solution `=(NxxExxV)/(1000)=(0.8xx53.5xx88.28)/(1000)=3.7783 g` [Eq. mass of `NH_(4)Cl=53.5]` |
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