Dissolving 120 g of urea (mol. wt. 60) in 10 | vedclass

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

Question

(C) Molarity is defined as the number of moles of solute dissolved in per litre of solution.Molarity,$M = \frac{n}{V}$$n =$ number of moles of solute

Vedclass · Accepted Answer

$2.05$

Vedclass · Answer

(C) Molarity is defined as the number of moles of solute dissolved in per litre of solution.Molarity,$M = \frac{n}{V}$$n =$ number of moles of solute = mass $/$ molar mass$V =$ volume of solutionMass of urea = $120 \ g$Molar mass of urea = $60 \ g / mol$Mass of water = $1000 \ g$Density of solution,$ho = 1.15 \ g / mL$$n = \frac{120}{60} = 2 \ mol$Total mass of solution = $m_{solute} + m_{solvent} = 120 \ g + 1000 \ g = 1120 \ g$Volume of solution,$V = \frac{	ext{mass}}{	ext{density}} = \frac{1120 \ g}{1.15 \ g / mL} \approx 973.91 \ mL = 0.97391 \ L$Molarity,$M = \frac{n}{V} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$Hence,the correct option is $C$.

Dissolving $120 \ g$ of urea (mol. wt. $60$) in $1000 \ g$ of water gave a solution of density $1.15 \ g / mL$. The molarity of the solution is (in $M$)

Similar Questions

You are supplied with $500 \ mL$ each of $2 \ N \ HCl$ and $5 \ N \ HCl$. What is the maximum volume of $3 \ M \ HCl$ that you can prepare using only these two solutions (in $mL$)?

$200 \, mL$ of $0.6 \, N$ $H_2SO_4$ and $100 \, mL$ of $0.3 \, N$ $HCl$ are mixed together. The acidic normality of the resultant solution is $........ N$.

The density of a solution containing $13 \%$ by mass of $H_2SO_4$ is $1.09 \ g/mL$. The molarity and normality of the solution are respectively :-

What volume of water in $mL$ is required to prepare a $0.20\,N$ solution from $1600\,mL$ of a $0.2050\,N$ solution?

What is the molarity of a solution containing $3.2 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $250 \ cm^{3}$ of water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module