A solution is prepared by adding 2 g of a su | vedclass

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

Question

(A) Mass per cent of solute $(A)$ is calculated using the formula:$	ext{Mass per cent} = \frac{	ext{Mass of solute}}{	ext{Mass of solution}} 	imes

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Mass per cent of solute $(A)$ is calculated using the formula:$	ext{Mass per cent} = \frac{	ext{Mass of solute}}{	ext{Mass of solution}} 	imes 100$Given:Mass of solute $(A)$ = $2 \ g$Mass of solvent (water) = $18 \ g$Mass of solution = $	ext{Mass of solute} + 	ext{Mass of solvent} = 2 \ g + 18 \ g = 20 \ g$$	ext{Mass per cent of } A = \frac{2 \ g}{20 \ g} 	imes 100 = 10 \%$

$A$ solution is prepared by adding $2 \ g$ of a substance $A$ to $18 \ g$ of water. Calculate the mass per cent of the solute. (in $\%$)

Similar Questions

$1 \, M$ solution contains

If $22 \ g$ of benzene $(C_6H_6)$ is dissolved in $122 \ g$ of carbon tetrachloride $(CCl_4)$,calculate the mass percentage of benzene. (in $\%$)

$A$ solution of $Al_2(SO_4)_3$ $(d = 1.253 \ g/mL)$ contains $22\%$ salt by weight. The molarity,normality,and molality of the solution are:

What type of solution is gasoline?

The density of a $1.0 \, m$ aqueous solution of glucose is $1.1 \, g/mL$. What is the molarity of the solution in $M$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module