(B) The number of moles of compound $A$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$. The v

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

(B) The number of moles of compound $A$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$. The volume of the solution in liters is: $V = 250 \ mL = 0.250 \ L$. The molarity $(M)$ is given by: $M = \frac{n}{V} = \frac{0.05 \ mol}{0.250 \ L} = 0.2 \ M$. Expressing this in the form $x \times 10^{-1}$: $0.2 = 2 \times 10^{-1}$. Therefore,the value of $x$ is $2$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

Medium

$4.5 \ g$ of compound $A$ $(MW = 90)$ was used to make $250 \ mL$ of its aqueous solution. The molarity of the solution in $M$ is $x \times 10^{-1}$. The value of $x$ is ............ (Rounded off to the nearest integer)

JEE MAIN 2021 All JEE MAIN

A
$1$
B
$2$
C
$3$
D
$4$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Method of expressing concentration of solution

Previous Year

JEE MAIN 2021 Paper All JEE MAIN years →

$A$ solution of a solute in benzene has a molality of $X \ m$. If the mole fraction of the solute is $0.2$,what is the value of $X$?

Medium

View Solution

$25 \ g$ of a solute of molar mass $250 \ g \ mol^{-1}$ is dissolved in $100 \ mL$ of water to obtain a solution whose density is $1.25 \ g \ mL^{-1}$. The molarity and molality of the solution are respectively

Medium

View Solution

The formula weight of $H_2SO_4$ is $98$. The weight of the acid in $400 \ mL$ of $0.1 \ M$ solution is ........... $g$.

Easy

View Solution

The density of $2 \ M \ MgBr_2$ solution is $1.2 \ g/mL$. Then the molality of the solution is ........ $m$.

Easy

View Solution

The density of an aqueous solution of $4\%$ by mass $NaOH$ is $1.04\ g/mL$. The molality of the solution is:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$4.5 \ g$ of compound $A$ $(MW = 90)$ was used to make $250 \ mL$ of its aqueous solution. The molarity of the solution in $M$ is $x \times 10^{-1}$. The value of $x$ is ............ (Rounded off to the nearest integer)

Similar Questions

$A$ solution of a solute in benzene has a molality of $X \ m$. If the mole fraction of the solute is $0.2$,what is the value of $X$?

$25 \ g$ of a solute of molar mass $250 \ g \ mol^{-1}$ is dissolved in $100 \ mL$ of water to obtain a solution whose density is $1.25 \ g \ mL^{-1}$. The molarity and molality of the solution are respectively

The formula weight of $H_2SO_4$ is $98$. The weight of the acid in $400 \ mL$ of $0.1 \ M$ solution is ........... $g$.

The density of $2 \ M \ MgBr_2$ solution is $1.2 \ g/mL$. Then the molality of the solution is ........ $m$.

The density of an aqueous solution of $4\%$ by mass $NaOH$ is $1.04\ g/mL$. The molality of the solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module