Calculate the molality of a solution containing 15.0 g of ethylene glycol (C2H6O2) dissolved in 145 g of water.

Molarity=moles of solute/ L of solution

Molality = moles of solute/ kg of solvent

Solute= what is being dissolved

Solvent= what is doing the dissolving

Solution= both together

Explanation:

Example's:

#1.  For number one you use the Molarity formula.  M= moles of solute/ L of solution.

To find moles of Mg(NO3)2 you divide 95g by its molar mass which is 148.33g so 95/148.33=.6405 moles of Mg(NO3)2.  Then plug in what you have.  .38M= .6405 moles Mg(NO3)2 / X.  Then solve for X using algebra.  .6405/.38= 1.686 L of solution.  (Volume).

Final 1.686 L

#2.  For number 2 you use the Molality formula.  m= moles of solute/ kg of solvent.

First you have to find moles of glucose by taking 267g and dividing it by its molar mass which is 180.56g.               267g/180.56g= 1.532 moles of glucose.  Then you have to change L to kg.  The easiest way to do this is to look at the density and see that for every 1 ml there is 1 gram.  So to take Liters to ml you multiply 1.59 by 1000 and get 1590 ml.  So that means you have 1590 grams. then you divide 1590grams by 1000 to get 1.59 Kg of slovent.  Then plug in your information into the formula.      molality= 1.532 moles of glucose / 1.59 Kg of solvent= .964 molality.

Final .964 mol/Kg

#3.  m= moles of solute / Kg of solvent.  0.445 mol solute / 2.07 Kg solvent= .215 Molality

Final .215 mol/Kg

#4. m= moles of solute / Kg of solvent.  take 13.5g and divide it by ethylene glycols molar mass which is 62.068 g.  13.5g / 62.068g= .218 mol.  Then you take 135g of water and divide it by 1000 to get Kg.  135/1000=.135 Kg.  Then plug in your information.  m= .218mol/.135 Kg= 1.615 molality

Final 1.615 mol/Kg.

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023