25 g of a solute of molar mass 250 g mol^{ | vedclass

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

Question

(D) $1$. Calculate moles of solute: $n = \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Calculate total mass of solution: $Mass_{solution} = Mas

Vedclass · Accepted Answer

$1 \ M$ and $1 \ m$

Vedclass · Answer

(D) $1$. Calculate moles of solute: $n = \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Calculate total mass of solution: $Mass_{solution} = Mass_{solute} + Mass_{solvent} = 25 \ g + 100 \ g = 125 \ g$. $3$. Calculate volume of solution: $V = \frac{Mass_{solution}}{Density} = \frac{125 \ g}{1.25 \ g \ mL^{-1}} = 100 \ mL = 0.1 \ L$. $4$. Calculate molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{0.1 \ L} = 1 \ M$. $5$. Calculate molality $(m)$: $m = \frac{n}{Mass_{solvent}(kg)} = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$. Thus,the molarity and molality are $1 \ M$ and $1 \ m$ respectively.

$25 \ g$ 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

Similar Questions

$9.8 \ g$ of $H_2SO_4$ is present in $2 \ L$ of a solution. The molarity of the solution is $....... \ M$

$8 \ g$ 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

Find out the mole fraction of glycerine in a solution containing $46 \ g$ of glycerine $(C_3H_8O_3)$ and $36 \ g$ of water $(H_2O)$.

Write the formula for percentage by weight.

What is the mole fraction of water in a spirit sample containing $92\%$ ethyl alcohol by mass?

Vedclass Test Series

Exam Paper Generator

Online Exam Module