Answer:
0.01 m Na3PO4 and 0.02 m KCl have the same boiling point with 0.04 m C6H12O6.
Explanation:
Okay, we are given 0.010 m Na3PO4 in water, 0.020 m CaBr2 in water, 0.020 m KCl in water, 0.020 m HF in water (HF is a weak acid.)
STEP ONE : write the equation of reaction for the dissociation of each compound given above.
Na3PO4 ------------> 3Na^+ + PO4^3-.
CaBr2 -----------------> Ca^2+ + 2Br^-.
KCl ----------------------------> K^+ + Cl^-.
HF -------------------------------> H^+ + F^-.
STEP TWO: from the dissociation reaction above we can get the Van't Holf factor,i.
For Na3PO4, Van't Holf factor,i= 4.
For CaBr2, Van't Holf factor,i = 3.
For KCl, Van't Holf factor,i = 2.
For HF, Van't Holf factor,i = 2.
STEP 3: we Calculate for the effective Concentration which can be calculated by;
Effective Concentration= Van't Holf factor,i × molarity.
For Na3PO4, Effective Concentration= 4 × 0.01 = 0.04.
For CaBr2, the Effective Concentration= 3 × 0.02 = 0.06.
For KCl, the effective Concentration= 2 × 0.02 = 0.04.
For HF, the effective Concentration= 2 × 0.02 = 0.04.
===>we can see that Na3PO4, KCl and HF has the same 0.04 Effective Concentration values with C6H12O6, then we can conclude that the three will be of the same boiling point as C6H12O6, BUT since HF is a weak acid and one of the properties of weak acid is that they dissociates partially. So, we will have to exclude HF out of the list. Therefore, only Na3PO4 and KCl have the same boiling point with C6H12O6.
