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

Calculate the molarity of ethanol in a solution where the mole fraction of ethanol is $0.040$. (Assume the density of water is $1 \ g/mL$.) (in $M$)

The concentration of $Ca^{2+}$ ions in a water sample is $0.0002 \ M$. What is the concentration of $Ca^{2+}$ in $ppm$ by weight-volume?

To what extent must a given solution of concentration of $40 \, mg$ silver nitrate per $mL$ be diluted to yield a solution of concentration of $16 \, mg$ $AgNO_3$ per $mL$?

$A$ solution is prepared by adding $2 \ g$ of $X$ to $1 \ mole$ of water. Mass percent of $X$ in the solution is $............. \ \%$

The density of a solution containing $13 \%$ by mass of $H_2SO_4$ is $1.09 \ g/mL$. The molarity and normality of the solution are respectively :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module