What is the concentration of sugar (C_{12}H_{ | vedclass

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

Question

(A) Molarity $(M)$ of a solution is defined as the number of moles of solute per liter of solution.$M = \frac{	ext{Number of moles of solute}}{	ext{

Vedclass · Accepted Answer

$0.02925 \ mol \ L^{-1}$

Vedclass · Answer

(A) Molarity $(M)$ of a solution is defined as the number of moles of solute per liter of solution.$M = \frac{	ext{Number of moles of solute}}{	ext{Volume of solution in Litres}}$First,calculate the molar mass of sugar $(C_{12}H_{22}O_{11})$:$Molar \ mass = (12 	imes 12) + (1 	imes 22) + (11 	imes 16) = 144 + 22 + 176 = 342 \ g \ mol^{-1}$Next,calculate the number of moles of sugar:$Moles = \frac{	ext{Mass}}{	ext{Molar mass}} = \frac{20 \ g}{342 \ g \ mol^{-1}} \approx 0.05848 \ mol$Finally,calculate the molarity:$M = \frac{0.05848 \ mol}{2 \ L} = 0.02924 \ mol \ L^{-1} \approx 0.02925 \ mol \ L^{-1}$

What is the concentration of sugar $(C_{12}H_{22}O_{11})$ in $mol \ L^{-1}$ if its $20 \ g$ are dissolved in enough water to make a final volume up to $2 \ L$?

Similar Questions

What is the molarity of $H_2SO_4$ solution if $25 \ mL$ is exactly neutralized with $32.63 \ mL$ of $0.164 \ M$ $NaOH$ (in $M$)?

How many grams of $60\% \, H_2SO_4$ are required for the complete reaction of $100 \, g \, CaCO_3$?

Which solution is required to completely neutralize $25 \ mL$ of $N/10$ $NaOH$ solution?

What is the volume of $CO_2$ in liters produced at $STP$ by the complete decomposition of $9.85 \ g$ of $BaCO_3$? (Atomic mass of $Ba = 137$,$C = 12$,$O = 16$)

$250 \ mL$ of a sodium carbonate solution contains $2.65 \ g$ of $Na_2CO_3$. If $10 \ mL$ of this solution is diluted to $1 \ L$, what is the concentration of the resultant solution (in $M$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module