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What is the change in the freezing point of water when 35.0 g of sucrose is
dissolved in 300.0 g of water?

Kf of water = -1.86°C/mol
molar mass sucrose = 342.30 g/mol
ivalue of sugar = 1

A. -0.183°C
B. 0.0606 °C
C. 0.217°C
D. -0.634°C

Respuesta :

pstnonsonjoku

From the calculations, we can see that, the change in the freezing point is -0.634°C.

What is freezing point?

The term freezing point refers to the temperature at which a liquid is changed to solid.

Given that;

ΔT = K  m i

Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles

molality =  0.1 moles/ 300.0 * 10^-3 Kg

= 0.33 m

Thus;

ΔT = -1.86°C/mol *  0.33 m * 1

= -0.634°C

Learn more about freezing point:https://brainly.com/question/3121416

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