A solution is 0.25 % by mass. What is the wei | vedclass

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

Question

(A) Mass percentage of solute $= 0.25 \%$. This means $0.25 \ g$ of solute is present in $100 \ g$ of solution.Mass of solvent $= 	ext{Mass of soluti

Vedclass · Accepted Answer

$498.75$

Vedclass · Answer

(A) Mass percentage of solute $= 0.25 \%$. This means $0.25 \ g$ of solute is present in $100 \ g$ of solution.Mass of solvent $= 	ext{Mass of solution} - 	ext{Mass of solute} = 100 \ g - 0.25 \ g = 99.75 \ g$.Thus,$0.25 \ g$ of solute is present in $99.75 \ g$ of solvent.For $1.25 \ g$ of solute,the mass of solvent required is $\frac{99.75 \ g 	ext{ solvent}}{0.25 \ g 	ext{ solute}} 	imes 1.25 \ g 	ext{ solute} = 498.75 \ g$.

$A$ solution is $0.25 \%$ by mass. What is the weight of solvent containing $1.25 \ g$ solute (in $g$)?

Similar Questions

The molarity of a $1 \ molal$ glucose solution having a density of $1.2 \ g/mL$ is: (in $M$)

What is the physical state of a solution containing camphor as the solute and $N_2$ gas as the solvent?

$A$ 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 $X$?

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

Which of the following modes of expressing concentration is independent of temperature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module