If equal weights of NaCl and KCl are dissolve | vedclass

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

Question

(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V} = \frac{w}{MW 	imes V}$.Since the weight $(w)$

Vedclass · Accepted Answer

Molarity of $NaCl$ is greater than $KCl$

Vedclass · Answer

(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V} = \frac{w}{MW 	imes V}$.Since the weight $(w)$ and volume $(V)$ are the same for both,$M \propto \frac{1}{MW}$.The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g/mol$.The molar mass of $KCl$ is $39 + 35.5 = 74.5 \ g/mol$.Since the molar mass of $NaCl$ $(58.5 \ g/mol)$ is less than the molar mass of $KCl$ $(74.5 \ g/mol)$,the molarity of the $NaCl$ solution will be greater than that of the $KCl$ solution.

If equal weights of $NaCl$ and $KCl$ are dissolved in the same volume of solution,how will the molarity of the two solutions compare?

Similar Questions

What is the mole fraction of the solute in a $1.00 \, m$ aqueous solution?

Molality of $30 \% \ w/w$ aqueous solution of $NaOH$ is . . . . . . . (in $m$)

What is meant by a binary solution?

Calculate the mass of Glucose $(C_6H_{12}O_6)$ required in making $2.5 \ kg$ of $0.25 \ molal$ aqueous solution. [Atomic wt : $H=1, O=16, C=12 \ amu$] (in $g$)

The molality and molarity of a solution of glucose in water which is labelled as $10 \% (w/w)$ are respectively (density of solution $= 1.2 \ g \ mL^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module