calculate the freezing point of a 0.09500 m aqueous solution of glucose. the molal freezing-point-depression constant of water is 1.86oc/m



Answer :

The freezing point of a 0.09500 m aqueous solution of glucose is -0.1767 °C.

Let's use the freezing point depression's accumulative feature.

ΔT=Kf.m.i

i= Van't Hoff factor (number of ions dissolved). Due to the non-electrolyte nature of glucose, I = 1

m = molality (mol of solute/kilogram of solvent).

We have these numbers → is 0.095 m

The freezing-point-depression constant for water Kf is 1.86 °C/m when

ΔT = T° freezing pure solvent - T° freezing solution

(0° - T° freezing solution)=  1.86 °C/m for . 0.095 m. 1

T° of freezing solution= 1.86 °C/m. 0.095 m . 1 → -0.1767°C

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