Answer:
(a) The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]
(b)The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]
Explanation:
The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]
Molality is expressed in units [tex]\frac{moles}{kg}[/tex]
(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 [tex]\frac{g}{mole}[/tex], then 14.3 grams of the compound represents the following number of moles:
[tex]14.3 grams*\frac{1 mole}{342 grams} =[/tex] 0.042 moles
Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:
[tex]molality=\frac{0.042 moles}{0.685 kg}[/tex]
Solving:
molality= 0.0613[tex]\frac{moles}{kg}[/tex]
The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]
(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:
[tex]molality=\frac{7.15 moles}{3.505 kg}[/tex]
Solving:
molality= 2.04[tex]\frac{moles}{kg}[/tex]
The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]
