How many moles and grams of sodium chloride a | vedclass

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

Question

(C) Given: Volume of solution $V = 200 \ cm^3 = 0.2 \ L$,Molarity $M = 0.2 \ M$.Formula: $	ext{Molarity} = \frac{	ext{Number of moles of solute}}{

Vedclass · Accepted Answer

$0.04, 2.34 \ g$

Vedclass · Answer

(C) Given: Volume of solution $V = 200 \ cm^3 = 0.2 \ L$,Molarity $M = 0.2 \ M$.Formula: $	ext{Molarity} = \frac{	ext{Number of moles of solute}}{	ext{Volume of solution in Liters}}$.Number of moles of $NaCl = 	ext{Molarity} 	imes 	ext{Volume in Liters} = 0.2 \ M 	imes 0.2 \ L = 0.04 \ 	ext{moles}$.Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.Mass of $NaCl = 	ext{Number of moles} 	imes 	ext{Molar mass} = 0.04 \ 	ext{mol} 	imes 58.5 \ g/mol = 2.34 \ g$.

How many moles and grams of sodium chloride are present in $200 \ cm^3$ of $0.2 \ M \ NaCl$ solution?

Similar Questions

The density of $NH_4OH$ solution is found to be $0.6 \, g/mL$. It contains $35 \%$ by mass of $NH_4OH$. The normality of the solution is ......... $N$.

Which of the following concentration factors is affected by a change in temperature?

Define the following terms:
$(i)$ Mole fraction $(ii)$ Molality $(iii)$ Molarity $(iv)$ Mass percentage

What are the mole fraction of water and the total number of moles in a solution formed by dissolving $1$ mole of urea in $1000 \ g$ of pure water,respectively?

If the mole fraction of the solute in liquid naphthalene $(C_{10}H_{8})$ is $0.2$,then the molality of the solution will be ............ $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module