A solution is prepared by dissolving 98.6 g of NaCI in enough water to form 875

Question

A solution is prepared by dissolving 98.6 g of NaCI in enough water to form 875 mL of solution. Calculate the mass % of the solution if the density of the solution is 1.06 g/mL A) 11.3% B) 12.7% C) 9.4% D) 10.6% E) 11.9% Place the following compounds in order of increasing strength of intermolecular forces. CH_4 CH_3CH_2CH_3 CH_3CH_3 A) CH_3CH_2CH_3 CO How much energy is required to vaporize 48.7 g of dichloromethane (CH_2CI_2) at its boiling its Delta H v_ap is 31.6 kJ/mol? A) 31.2 KJ b) 6.49 KJ C) 55.1 KJ D) 15.4 KJ

Explanation / Answer

7. density = mass / volume
Therefore mass = density x volume
= 1.06 g/ml x 875 ml
= 927.5 g

% mass NaCl = 98.6 g / 927.5 g x 100/1
= 10.6 % (d)

8. Larger alkanes-in particular, those that are longer, less branched chains-have more possible locations in which dipole moments can be induced, meaning that the van der Waals forces between these larger molecules can accumulate. So, intermolecular forces increasing trend is E. CH4<CH3CH3<CH3CH2CH3

9.c.

11. Under standard conditions, water is a liquid while N2 and CO are gases. Thus, water has the highest boiling point.

N2 is a liquid with a temperature of approx. 70K - thus it's probably got the lowest boiling point.

But it comes down to molecular forces between the molecules. N2 molecules have no preference - they can be near each other or all spread out. H2O molecules hydrogen bond and want to be closer to each other - it requires more energy to break this bond than the break the non-existant bonds between molecules of N2. CO is slightly polar - the molecules hold on and then eventually give it.

N2 - 77K
CO - 81K
H2O (pure) - 373K

Hence, answer is c) H2O > CO>N2

12. 48.7 g of CH2Cl2 = 48.7/86 mol of CH2Cl2 = 0.566 mol

delta H(vap) = 31.6 kJ/mol i.e. to vaporize 1 mol you need 31.6 kJ energy.

So, energy required to vaporize 0.566 mol of CH2Cl2 = 31.6 kJ/mol * 0.566 mol =17.7 kJ (E)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.