Calculate the freezing point of a 0.09500 m aqueous solution of glucose. the molal freezing-point-depression constant of water is is 1.86 *c/m

(a)-0.177
(b) 0.0475
(c) 0.106
(d)-0.0562
(e) -0.354

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

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

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

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

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms