1. Common commercial hydrochloric acid is an aqueous solution that is 38% HCl by

Question

1. Common commercial hydrochloric acid is an aqueous solution that is 38% HCl by mass, with a density of 1.19 g/cm^3. Calculate the molarity of this reagent. Molarity = ___mol/L Calculate the molality of this reagent. Molality = ___mol/kg Calculate the mole fraction of this reagent. Mole fraction = 2. Common commercial nitric acid is an aqueous solution that is 70.% HNO3 by mass, with a density of 1.42 g/cm^3. Calculate the molarity of this reagent. Molarity = ___mol/L Calculate the molality of this reagent. Molality = ___mol/kg Calculate the mole fraction of this reagent. Mole fraction = 3. Common commercial sulfuric acid is an aqueous solution that is 95% H2SO4 by mass, with a density of 1.84 g/cm^3. Calculate the molarity of this reagent. Molarity = ___mol/L Calculate the molality of this reagent. Molality = ___mol/kg Calculate the mole fraction of this reagent. Mole fraction = 4.Common commercial acetic acid is an aqueous solution that is 99% by mass CH3CO2H, with a density of 1.05 g/cm^3. Calculate the molarity of this reagent. Molarity = ___mol/L Calculate the molality of this reagent. Molality = ___mol/kg Calculate the mole fraction of this reagent. Mole fraction = 5. Common commercial ammonia is an aqueous solution that is 28% NH3 by mass, with a density of 0.90 g/cm^3. Calculate the molarity of this reagent. Molarity = ___mol/L Calculate the molality of this reagent. Molality = ___mol/kg Calculate the mole fraction of this reagent. Mole fraction =

Explanation / Answer

1) HCl

Given density = 1.19 g /cm^3

Mass% = 38 g HCl / 100g of solution

So mass of 38 g is in 100 / 1.19 cm^3 of solution

Or 38 g in 84.03 mL

So mass in 1L = 452.2 grams

Moles of Hcl = 452.2 / 36.5 = 12.39

So molarity = 12.39 moles / L

Mass of water in 100g = 100 - 38 = 62 grams

So 38 grams in 62 grams of water

So mass of HCl in 1000grams of water = 38 * x 1000 / 62 = 612.9

Moles of HCl in 1000g = 612.9 / 36.5 = 16.79 moles
Molality = 16.79 molal (mole / Kg)

Mole fraction = Mole of HCl / Moles of water + mole of HCl

Mole of water = 62/ 18 = 3.44

Moles of Hcl = 38 / 36.5 = 1.032

Mole fraction = 1.032 / 3.44 + 1.032 = 0.23

2) Again we will use same approach

Mass of HNO3 in 100g =70 grams

Volume of solution= 100 / 1.42 mL = 70.42 mL

So mass of HNO3 in 70.42 mL = 70 grams

Mass of HNO3 in 1000mL (1L) = 70X 1000 / 70.42 = 994.03 grams

Moles = 994.03 / 63 = 15.78 moles

So molarity = 15.78 moles / L

b) Mass of HNO3 is 70grams in 30 grams of water

So Mass of HNO3 in 1000grams (1Kg) of water = 70*1000/ 30 = 2333.3 grams

Moles ofHNO3 in 1000grmas of water = 2333.3 / 63 = 37.04 moles

So molality = Moles / Kg of solvent = 37.04 Molal

c) moles of HNO3 in 100grams = 70 / 63 = 1.11

Moles of Water = 30 / 18 = 1.67

So mole fraction of HNO3 = 1.11 / 1.67 + 1.11 = 0.399

3) Again we will use same approach

Mass of H2SO4 in 100g =95 grams

Volume of solution= 100 /1.84 mL = 50.35 mL

So mass of H2SO4 in 50.35 mL = 95 grams

Mass of H2SO4 in 1000mL (1L) =95 X 1000 / 50.35 = 1886.79 grams

Moles = 1886.79 / 98 = 19.25 moles

So molarity = 19.25 moles / L

b) Mass of H2SO4 is 95grams in 5 grams of water

So Mass of H2SO4 in 1000grams (1Kg) of water = 95*1000/ 5 =19000 grams

Moles of H2SO4 in 1000grmas of water = 19000 / 98 =193.87 moles

So molality = Moles / Kg of solvent =193.87 Molal

c) moles of H2SO4 in 100grams = 95 / 98 = 0.969

Moles of Water = 5 / 18 = 0.278

So mole fraction of HNO3 = 0.969 / 1.247 = 0.777

4) Again we will use same approach

Mass of CH3COOH in 100g =99 grams

Volume of solution= 100 /1.05 mL = 95.23 mL

So mass of CH3COOH in 95.23 mL = 99 grams

Mass of CH3COOH in 1000mL (1L) =99 X 1000 / 95.23 = 1039.58 grams

Moles = 1039.58 / 60 = 17.32 moles

So molarity = 17.32 moles / L

b) Mass of CH3COOH is 95 grams in 5 grams of water

So Mass of CH3COOH in 1000grams (1Kg) of water = 99*1000/ 1 =99000 grams

Moles of CH3COOH in 1000grmas of water = 99000 / 60 =1650 moles

So molality = Moles / Kg of solvent =1650 Molal

c) moles ofCH3COOH in 100grams = 99/ 60 = 1.65

Moles of Water = 1 / 18 =0.05

So mole fraction of CH3COOH = 1.65 / 170 = 0.97

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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