Asolution is made by dissolving 23.5 grams of glucose (c6h12o6) in 0.245 kilograms of water. if the molal freezing point constant for water (kf) is -1.86 °c/m, what is the resulting δf of the solution? show all the steps taken to solve this problem. (4 points)

- 0.99 °C ≅ - 1.0 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(0.53 m) = - 0.99 °C ≅ - 1.0 °C.

Mass is the amount of matter in an object.  weight is the measurement of the pull of gravity on an object.  say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165)  now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon.  however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant.  in conclusion it is possible for your weight to change as your mass remains constant. 

You would do 1500/4 and the answer is 375