Sucrose (C 12 ,H 22 ,O 11 ), a nonionic solute, dissolves in water (normal freez

Question

Sucrose (C12,H22,O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0° C) to form a solution. If some unknown mass of sucrose is dissolved in 150g of water and this solution has a freezing/melting point of -.56° C, calculatue the mass of sucrose dissolved. Kfp for water is 1.86°C/m. YOU MUST USE THE FOLLOWING 2 EQUATIONS AND SHOW WORK.

Tfp= iKfpm solute where Tfp = Ti - Tf and is positive

Tfp = (Wsolute/(MsoluteWSolvent))Kfp

Note: I do not know how to solve this, but you may only need to use the second equation. Thank you!

Explanation / Answer

We know that T f = Kf x m
Where
T f = depression in freezing point
= freezing point of pure solvent – freezing point of solution
= 0-(-56)= 56 oC
K f = depression in freezing constant = 1.86 oC/ m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg

= (m/342(g/mol))/0.15 kg

= 0.02 m

Plug the values we get m = 1505 g

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