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

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(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 ........... $molal$.

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