10 mL of 2 M NaOH solution is added to 20 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl ightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M 	imes 0.01 \ L = 0

Vedclass · Accepted Answer

$20 \ M$ $HCl$ solution

Vedclass · Answer

(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker: $NaOH + HCl ightarrow NaCl + H_2O$ Initial moles of $NaOH = 2 \ M 	imes 0.01 \ L = 0.02 \ mol$. Initial moles of $HCl = 1 \ M 	imes 0.02 \ L = 0.02 \ mol$. Since the moles are equal,the resulting solution is neutral ($NaCl$ solution). Step $2$: $10 \ mL$ of this neutral solution is added to a flask containing $2 \ moles$ of $HCl$. The total volume of the flask is made up to $100 \ mL$ $(0.1 \ L)$. Molarity of $HCl = \frac{	ext{moles of } HCl}{	ext{Volume of solution in } L} = \frac{2 \ mol}{0.1 \ L} = 20 \ M$.

Similar Questions

What volume of $NH_3$ gas at $STP$ would be needed to prepare $100 \ mL$ of $2.5$ molal $(2.5 \ m)$ ammonium hydroxide solution? (Assume the density of the solution is $1 \ g/mL$)

Which is the correct option for $0.1 \ M$,$500 \ mL$ of $AgCl$?

For the stoichiometric reaction occurring as shown in the figure,which option is incorrect?

Which sample contains the largest number of atoms?

When a standard solution of $NaOH$ is left in air for a few hours:

Vedclass Test Series

Exam Paper Generator

Online Exam Module