An aqueous solution has a mole fraction of 0.072 koh. what is the molarity of the solution? assume that the density of the solution is 1.15 g/ml. enter your answer to three significant figures and in units of molarity.

The molarity of the solution is 3.99 M

Explanation:

Step 1: Data given

Mole fraction = 0.072 KOH

Density of the solution = 1.15 g/mL

Step 2: Calculate mass of the solution

1000 mL *  1.15 g/mL  = 1150 grams solution

Step 3: Calculate moles of KOH

moles KOH/  (moles KOH  +  moles H2O) = 0.072

Suppose x = grams KOH

Moles of KOH = x/56g/mol

moles H2O = (1150 -x)/18

1150 grams solution – grams KOH = grams of H2O  

(x/56) / (x/56 + (1150 –x)/18) = 0.072

0.017857x / (0.017857x  +  63.89 – 0.0556x) = 0.072

x =  223.44 grams KOH

Moles of KOH = 223.44 grams / 56g/mol = 3.99 moles KOH

Step 4: Calculate molarity of the solution  

223.44  grams KOH / 1 liter *  1 mole KOH/ 56 g/mol =  3.99 mol/L

Molarity =  3.99 M

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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.