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 does the equation $2Al(s) + \frac{3}{2}O_2(g) \to Al_2O_3(s)$ represent?

The volume of $O_2$ required for the complete combustion of $5$ moles of propane $(C_3H_8)$ at $25$ $^\circ C$ is ............... liters.

The reaction of $1 \ mol$ of magnesium nitride with excess water gives:

At $T \ K$,$3 \ moles$ of hydrogen and $1 \ mole$ of $N_2$ are allowed to react to form ammonia. When $1 \ mole$ of ammonia is formed,the total pressure in the vessel is $15 \ atm$. The partial pressure of $N_2$ in the vessel (in $atm$) is (in $.5$)

The normality of $1 \ M$ solution of $H_3PO_4$ will be.....$N$

Vedclass Test Series

Exam Paper Generator

Online Exam Module