[nawinince]

(1) 9.0 g of C6H12O6, was dissolved in 100 g of water. The freezing point of the glucose solution was determined to be -0.94°C. Determine the molar freezing-point depression of water in the unit of (K kg/mol)

my attempt :

molal = (9/180)/100*10^-3 = 1/2

T=k*molal

0.94 = K*1/2
K=1.88

but the answer is 1.9

what did i miss?

[sjb]

Nothing really (sig fig? units?)

[nawinince]

Nothing really (sig fig? units?)

So my solution was correct? if so, my question must has mistake.

[Borek]

So my solution was correct? if so, my question must has mistake.

Your calculations are OK, you just reported too many significant digits (you were expected to round down the 1.88 to 1.9).

https://en.wikipedia.org/wiki/Significant_figures