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A solution was prepared by dissolving 27.1g of sucrose (C12H22O11) in 250 g of water. Find the molality of the solution(molar mass of C12H22O11 is 342.8)
-.108 m
-108 m
-317 m
-.317 m

Respuesta :

ajeigbeibraheem

Answer:

M = 0.317 m

Explanation:

From the given information:

Using the following formula:, the molality M is:

[tex]M= \dfrac{W_A}{M_A}\times \dfrac{1000}{W_B(g)}[/tex]

where;

M = molality

[tex]W_A[/tex]= mass of the solute = 27.1 g

[tex]M_A[/tex] = molar mass of the solute = 342.8

[tex]W_B[/tex] = mass of the given solvent  = 250 g

∴

[tex]M= \dfrac{W_A}{M_A}\times \dfrac{1000}{W_B(g)}[/tex]

[tex]M= \dfrac{27.1}{342.8}\times \dfrac{1000}{250}[/tex]

M = 0.317 m