Question

The $M$ solution required for $NaOH$ in $40.50mL$ from solution containing $2g$ of $NaOH$ in $500mL$ for complete neutralisation is:

• A. $10mL$ decinormal $HCl$
• B. $20mL$ decinormal $HCl$
• C. $50mL$ decinormal $HCl$
• D. $25mL$ decinormal $HCl$

Answer 1

Answer: (C)

$50mL$ decinormal $HCl$

molecular mass of NaOH = 40 g

molarity of stock solution = $\frac{2}{40\times 500}\times 1000$ = 0.1 M

no. of moles of NaOH in 40.50 ml of solution = $0.1\times \frac{40.5}{1000}$ = 0.004504 moles

for neutralisation, we have: no. of moles of HCl = no. of moles of NaOH

1 decimolar = 0.1 M

so, number of moles of HCl in 10 ml decimolar solution = $\frac{10}{1000}\times 0.1$ = 0.001

number of moles of HCl in 20 ml decimolar solution = $\frac{20}{1000}\times 0.1$ = 0.002

number of moles of HCl in 50 ml decimolar solution = $\frac{50}{1000}\times 0.1$ = 0.005