A solution is obtained by mixing 400 g of 25 | vedclass

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

Question

(C) Mass of solute in the first solution $= 400 \ g 	imes 0.25 = 100 \ g$. Mass of solute in the second solution $= 600 \ g 	imes 0.40 = 240 \ g$. T

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(C) Mass of solute in the first solution $= 400 \ g 	imes 0.25 = 100 \ g$. Mass of solute in the second solution $= 600 \ g 	imes 0.40 = 240 \ g$. Total mass of solute $= 100 \ g + 240 \ g = 340 \ g$. Total mass of the resulting solution $= 400 \ g + 600 \ g = 1000 \ g$. Mass percentage of the resulting solution $= (	ext{Total mass of solute} / 	ext{Total mass of solution}) 	imes 100 = (340 \ g / 1000 \ g) 	imes 100 = 34 \%$. Therefore,the correct option is $C$.

$A$ solution is obtained by mixing $400 \ g$ of $25 \%$ solution and $600 \ g$ of $40 \%$ solution by mass. What is the mass percentage of the resulting solution (in $\%$)?

Similar Questions

If $5.85 \ g$ of $NaCl$ is dissolved in $90 \ g$ of water,then find out the mole fraction of $NaCl$.

$8 \ g$ of $NaOH$ is dissolved in $18 \ g$ of $H_2O$. Mole fraction of $NaOH$ in solution and molality (in $mol \ kg^{-1}$) of the solution respectively are

What is meant by mass percentage $( \% w/w )$?

If the mole fraction of the solute in benzene is $0.2$,then the molality of the solution will be ..............

Which of the following properties will change when the system containing solution $1$ becomes solution $2$?

$10 \ mol$ of solute $x + 10 \ L$ of water $1 \ L$ of solution $1 + 1 \ mol$ of solute $x + 1 \ L$ of water

(Solution $1$) (Solution $2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module