The molecular weight of glucose is 180. A sol | vedclass

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

Question

(C) The molarity $(M)$ of the solution is calculated as: $M = \frac{	ext{mass of solute}}{	ext{molar mass} 	imes 	ext{volume of solution in litres

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(C) The molarity $(M)$ of the solution is calculated as: $M = \frac{	ext{mass of solute}}{	ext{molar mass} 	imes 	ext{volume of solution in litres}} = \frac{18 \ g}{180 \ g/mol 	imes 1 \ L} = 0.1 \ M$. Assuming the density of the solution is approximately $1 \ g/mL$,the mass of $1 \ L$ of solution is $1000 \ g$. The mass of the solvent (water) is $1000 \ g - 18 \ g = 982 \ g = 0.982 \ kg$. Molality $(m)$ is defined as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \ mol}{0.982 \ kg} \approx 0.102 \ molal$. Given the options provided and the context of such problems,the intended answer is $0.1$.

The molecular weight of glucose is $180$. $A$ solution of glucose which contains $18 \ g$ per litre is ........... $molal$.

Similar Questions

Two solutions labelled as $3 \, M \, HCl$ and $1 \, M \, HCl$ are mixed in the ratio of $x : y$ by volume and the molarity of the resulting solution is $1.5 \, M$. What is the molarity of the resulting solution if they are mixed in the ratio $y : x$ by volume?

The density of $3 \ m$ solution of $NaOH$ is $1.110 \ g \ mL^{-1}$. Calculate the molarity of the solution. (in $M$)

Calculate the concentration of sugar $(C_{12}H_{22}O_{11})$ in $mol \ L^{-1}$ if $20 \ g$ of sugar is dissolved in sufficient water to make the final volume $2 \ L$.

The molality of $15\% \,(w/V)$ solution of $H_2SO_4$ with density $1.1\, g/cm^3$ is

The molality of an aqueous solution of glucose is $1.0 \ m$. The mole fraction of glucose is approximately .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module