What is the molarity of a solution prepared b | vedclass

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

Question

(D) To calculate the molarity $(M)$,we use the formula: $M = \frac{	ext{moles of solute}}{	ext{volume of solution in liters}}$.First,calculate the m

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) To calculate the molarity $(M)$,we use the formula: $M = \frac{	ext{moles of solute}}{	ext{volume of solution in liters}}$.First,calculate the molar mass of $NaOH$: $Na = 23, O = 16, H = 1$. So,$23 + 16 + 1 = 40 \ g/mol$.Next,calculate the number of moles of $NaOH$: $	ext{moles} = \frac{	ext{given mass}}{	ext{molar mass}} = \frac{8 \ g}{40 \ g/mol} = 0.2 \ mol$.Convert the volume from $mL$ to $L$: $100 \ mL = 0.1 \ L$.Finally,calculate molarity: $M = \frac{0.2 \ mol}{0.1 \ L} = 2 \ M$.Therefore,the correct option is $D$.

What is the molarity of a solution prepared by dissolving $8 \ g$ of $NaOH$ in water to make a total volume of $100 \ mL$ (in $M$)?

Similar Questions

The Bronsted-Lowry acid-base theory was based on which of the following?

The compound which is used as an oxidising agent in many chemical industries is:

Which of the following solutions is most basic?

What is the $pH$ value of a basic aqueous solution?

Explain why:
$(a)$ the colour of copper sulphate crystals changes on heating?
$(b)$ baking soda acts as an antacid?
$(c)$ an acid should be added to water while diluting?

Vedclass Test Series

Exam Paper Generator

Online Exam Module