Question

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

Answer 1

M = 0.317 m

Step-by-step explanation:

From the given information:

Using the following formula:, the molality M is:

`M= (W_A)/(M_A)* (1000)/(W_B(g))`

where;

M = molality

`W_A`= mass of the solute = 27.1 g

`M_A` = molar mass of the solute = 342.8

`W_B` = mass of the given solvent = 250 g

∴

`M= (W_A)/(M_A)* (1000)/(W_B(g))`

`M= (27.1)/(342.8)* (1000)/(250)`

M = 0.317 m