If 18 g of glucose (C_6H_{12}O_6) is present | vedclass

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

Question

(D) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$. Given mass of glucose = $18 \ g$. Moles of glucose = $\frac{18 \ g}{180 \ g/mol} = 0.

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(D) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$. Given mass of glucose = $18 \ g$. Moles of glucose = $\frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$. Mass of solvent (water) = $1000 \ g - 18 \ g = 982 \ g = 0.982 \ kg$. Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \ mol}{0.982 \ kg} \approx 0.1018 \ m$. Since the question implies a standard approximation where the mass of the solvent is taken as $1000 \ g$ $(1 \ kg)$ for dilute solutions,the molality is $0.1 \ m$.

If $18 \ g$ of glucose $(C_6H_{12}O_6)$ is present in $1000 \ g$ of an aqueous solution of glucose,it is said to be ......... $molal$.

Similar Questions

If a $3.6 \, M$ solution of $H_2SO_4$ contains $29 \% \, H_2SO_4$ by mass,find the density of the $H_2SO_4$ solution. (Molar mass of $H_2SO_4 = 98 \, g \, mol^{-1}$)

The unit of molality is

$3.65 \ g$ of $HCl$ is dissolved in $16.2 \ g$ of water. The mole fraction of $HCl$ in the resulting solution is ......

$A$ sample of drinking water is severely contaminated with chloroform $(CHCl_3)$. $CHCl_3$ is considered carcinogenic in nature. The level of contamination was $15 \ ppm$ (by mass).
$(i)$ Express this in percent by mass.
$(ii)$ Determine the molality of chloroform in the water sample.

If the density of a solution is $1.17 \, g/cc$,what is the molarity of liquid $HCl$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module