1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxi

Question

1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

Professor PC Thomas Classes · Answer

(b) : Molarity $=\frac{	ext { Moles }}{	ext { Volume }(\mathrm{mL})} 	imes 1000$ For HCl , No. of moles $=\frac{M 	imes V(\mathrm{~mL})}{1000}=\frac{0.75 	imes 25}{1000}=\frac{0.75}{40}$ \ Mass of HCl reacted $=\frac{0.75}{40} 	imes 36.5=0.684 \mathrm{~g}$  \ 36.5 g HCl reacts with 40 g NaOH
0.684 g HCl will react with $\frac{40}{36.5} 	imes 0.684=0.75 \mathrm{~g}$
NaOH left unreacted $=1-0.75=0.25 \mathrm{~g}$ or 250 mg