Which lowers the freezing point of 2.0 kg of water more, 2.0 mol NaOH or 2.0 mol

Question

Which lowers the freezing point of 2.0 kg of water more, 2.0 mol NaOH or 2.0 mol Ba(OH)2? (Both compounds are strong electrolytes and completely dissociate into ions when dissolved.) Explain in 1-2 sentences. Dilute solutions of NaHCO3 are sometimes used in treating acid bums. How many milliliters of a 0.100 M NaHCO3 solution are needed to prepare 750.0 mL of 0.0500 M NaHCO3 solution? How many milliliters of a 0.75 M HC1 solution do you need to obtain 0.0040 mol of HC1? Why do red blood cells swell up and burst when placed in pure water? Explain in 1-2 sentences. How many moles of solute are present in 480 mL of 1.4 M HNO3? Extra Credit: Describe how you would prepare 500 mL of a 0.20 M NaCl solution (starting from solid NaCl).

Explanation / Answer

Solution :-

Q1) 2.0 mol NaOH will give 2.0 mol * 2 ions / mol = 4.0 mol ions

2.0mol Ba(OH)2 will give 2.0 mol * 3 ions / mol = 6.0 mol ions

The freezing point of the solution is depends on the total ion concentration therefore

2.0mol Ba(OH)2 will decrease the freezing point of 2.0 kg water more.

So the answer is 2.0 mol Ba(OH)2

Q2) initial molarity M1 = 0.100 M NaHCO3

Initial volume V1 = ?

Final volume V2 = 750.0 ml

Final molarity M2 = 0.0500 M

Using the formula M1V1=M2V2 we can calculate the initial volume of the solution needed to make the dilute solution

V1 = M2V2/M1

      = 0.0500 M * 750.0 ml / 0.100 M

     = 375.0 ml

So the volume of the 0.100 M solution needed is 375.0 ml

Q3 )

Moles = molarity * volume in liter

Therefore

Volume in liter = moles / molarity

                            = 0.0040 mol HCl / 0.75 mol per L

                           =0.00533 L

Lets convert liter to ml

0.00533 L * 1000 ml / 1 L = 5.33 ml

So the volume of the HCl needed is 5.33 ml

Q4) When the red blood cells are placed in the pure water then concentration of the liquid inside the cell is higher therefore the water enters the cell to dilute the solution but this increase volume of the solution creates the pressure on the cell wall therefore they burst.

Q5) 1.4 M HNO3 , volume = 480 ml

480 ml = 0.480 L

Moles of solute = molarity * volume  

                          = 1.4 mol per L *0.480 L

                      = 0.672 mol Solute

Extra credit

500 ml = 0.500 L

0.20 M NaCl

Moles of NaCl = molarity * volume  

                          = 0.20 mol per L *0.500 L

                          = 0.1 mol NaCl

Mass of NaCl = moles * molar mass

                         = 0.100 mol * 58.443 g per mol

                        = 5.84 g NaCl

So need to use 5.84 g NaCl and dissolve in water and dilute it to total volume of 500 ml

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