Asolution was prepared by dissolving glucose (c6h12o6, mm 180) in 100.0 grams of acetic acid. the boiling point of the solution was found to be 119.81 c. the beoiling point of acetic acid is 118.10 c and the boiling point elevation constant is 3.07 c/m. how many grams of glucose were dissolved? a) 10.0gb) 15.1gc) 21.0gd) 17.8ge) 21.0g

a) 10.0 g

Explanation:

The glucose is a nonvolatile compound, so when added to a solvent, it will change the boiling point of this substance, which is called ebullioscopy. The change in temperature can be calculated by:

ΔT = K*W*i

Where K is the constant, W is the molality, and i is the van't Hoff factor.

The van't Hoff factor is related to the dissociation of the compound. For molecular compounds, such as glucose, i = 1.

119.81 - 118.10 = 3.07*W

3.07W = 1.71

W = 0.5570 mol/kg

The molality can be calculated by:

W = m1/M1*m2

Where m1 is the mass of the solute in g, M1 is the molar mass of the solute, and m2 is the mass of the solvent in kg.

0.5570 = m1/180*0.1

m1 = 10.026 g ≅ 10.0 g

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explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.