Estimate the freezing point depression of 0.19 m C12H22O11(aq). The freezing point constant of water is 1.86 K · kg/mol.
Answer in units of ◦C.

0.3634 °C

Explanation:

In this question, we are tasked with calculating the freezing point depression of 0.19 M C12H12O11( sucrose)

Mathematically this can be calculated using the formula;

ΔTf = i * kf * m

where ΔTf is the freezing point depression

i is Van’t Hoff factor

kf is the molal freezing point depression for the solvent

m is the molality of the solution

From the question we identify the following;

Since sucrose is a non-electrolyte, i = 1 ( i refers to the number of moles of particles obtained when 1 mol of solute dissolves)

m is 0.19

kf = 1.86

substituting these values into the equation, we have

ΔTf = 1 * 1.86 * 0.19 = 0.3534 °C

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72