60 g of CH_3COOH is dissolved in 1 dm^3 of | vedclass

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

Question

(A) Given: Mass of solute $(CH_3COOH)$ $= 60 \ g$,Volume of solution $= 1 \ dm^3 = 1000 \ cm^3$,Density of solution $= 1.25 \ g / cm^3$. Molar mass of

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(A) Given: Mass of solute $(CH_3COOH)$ $= 60 \ g$,Volume of solution $= 1 \ dm^3 = 1000 \ cm^3$,Density of solution $= 1.25 \ g / cm^3$. Molar mass of $CH_3COOH = 12 + 3 + 12 + 16 + 16 + 1 = 60 \ g / mol$. Moles of solute $= \frac{60 \ g}{60 \ g / mol} = 1 \ mol$. Mass of solution $= 	ext{Volume} 	imes 	ext{Density} = 1000 \ cm^3 	imes 1.25 \ g / cm^3 = 1250 \ g$. Mass of solvent $= 	ext{Mass of solution} - 	ext{Mass of solute} = 1250 \ g - 60 \ g = 1190 \ g = 1.19 \ kg$. Molality $(m) = \frac{	ext{Moles of solute}}{	ext{Mass of solvent in kg}} = \frac{1 \ mol}{1.19 \ kg} \approx 0.84 \ m$. Note: Assuming the $1 \ dm^3$ refers to the volume of the solution,the calculated molality is approximately $0.84 \ m$. Given the options,$0.8 \ m$ is the closest approximation.

$60 \ g$ of $CH_3COOH$ is dissolved in $1 \ dm^3$ of solvent. What is the molality of the solution (in $m$)? (Density $= 1.25 \ g / cm^3$)

Similar Questions

What is the mole fraction of water in $10\%$ by weight $(w/w)$ of aqueous urea solution? [Given: Molar mass of $H, O, C$ and $N$ are $1, 16, 12$ and $14 \text{ g mol}^{-1}$ respectively.]

The volume of water added to $500 \ mL$ of $0.5 \ M \ NaOH$ solution so that its strength becomes $10 \ mg \ mL^{-1}$ will be .......... $mL$.

Which of the following concentration factors changes with temperature?

An aqueous solution of $98\% (w/w)$ $H_2SO_4$ has a density of $1.84 \ g/mL$. The molality of the solution is: (in $m$)

The molality of a solution containing $15.20 \ g$ of urea (molar mass = $60 \ g \ mol^{-1}$) dissolved in $150 \ g$ of water is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module