Which has the lowest freezing point? a. pure water b. 1.0 m nacl c. 1.0 m k3po4 d. 1.0 m cacl2 e. 1.0 m glucose (c6h12o6)?

Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures. 
2) The more particles of the solute in the solution the lower freezing point is going to be.
b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)
d. 1.0 m CaCl2 (dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))

The largest number of particles has 1.0 m K3PO4  solution, and it is has lowest freezing point . Answer is C.

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³

Heat is how much heat energy something has and temperature is a measure of how hot something is.