Question

part a.) A student needs 155 mL of 2.9M HCl solution for an experiment. What volume (in mL) of 10M HCl would need to be diluted to make the desired solution?

part b.) Calcium carbonate, also known as limestone, ia decomposed by heating into calcium oxide and carbon dioxide. A sample of calcium carbonate is decomposed and the CO2 is collected in a 0.15L flask at 72 degrees celcius, the CO2 collected was found to have a pressure of 1.3 atm. How many moles of CO2 were collected?

part c.) what would be the new boiling point of a solution made from 2.3 moles of NaCl and 1.1kg of water?

part d.) a solution is made by dissolving 144g of NaCl in 954mL of water. what is the molality of the solution?

Answer 1

part a.) A student needs 155 mL of 2.9M HCl solution for an experiment. What volume (in mL) of 10M HCl would need to be diluted to make the desired solution?

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

M1V1 = M2V2

155*2.9 = 10*V

V = 155*2.9/10

V = 44.95 mL or stock solution

part b.) Calcium carbonate, also known as limestone, ia decomposed by heating into calcium oxide and carbon dioxide. A sample of calcium carbonate is decomposed and the CO2 is collected in a 0.15L flask at 72 degrees celcius, the CO2 collected was found to have a pressure of 1.3 atm. How many moles of CO2 were collected?

CaCO3 = CaO + CO2

PV = nRT

n = PV/(RT) = (1.3*0.15)/(0.082*(72+273)) = 0.006892 mol of CO2

part c.) what would be the new boiling point of a solution made from 2.3 moles of NaCl and 1.1kg of water?

dTb = Kb*m*i

m = molality = mol of NaCl / Kg of water = 2.3/1.1 = 2.09

Kb = 0.512; I = 2 ions

dTb = 0.512*2.09 *2= 2.14016

Tb = 100 + 2.14016= 102.14016 °C

part d.) a solution is made by dissolving 144g of NaCl in 954mL of water. what is the molality of the solution?

kg of water = 0.954 kg, mol of NACl = mass/ W = 144/55.8 = 2.580

molality = mol of NaCl / kg ofwater = 2.580/0.954 = 2.704