Concentrated hydrochloric acid solution contains 37.0% by mass HCl. If the density is 1.18 g/mL, what is the molar concentration? (LO 13.6, 13.7) (a) 10.1 M (b) 16.2 M (c) 8.60 M (d) 12.0 M

Verified step by step guidance

Step 1: Determine the mass of HCl in 1 liter of solution. Since the solution is 37.0% HCl by mass, this means that in 100 g of solution, there are 37.0 g of HCl.

Step 2: Calculate the mass of 1 liter of the solution using its density. The density is given as 1.18 g/mL, so 1 liter (1000 mL) of solution has a mass of 1.18 g/mL * 1000 mL = 1180 g.

Step 3: Calculate the mass of HCl in 1 liter of solution. Using the percentage by mass, the mass of HCl in 1180 g of solution is (37.0/100) * 1180 g.

Step 4: Convert the mass of HCl to moles. Use the molar mass of HCl, which is approximately 36.46 g/mol, to convert the mass of HCl to moles.

Step 5: Calculate the molar concentration (molarity) of the solution. Molarity is defined as moles of solute per liter of solution. Use the moles of HCl calculated in Step 4 and divide by the volume of the solution in liters (1 L) to find the molarity.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mass Percent Concentration

Mass percent concentration is a way to express the concentration of a solution, defined as the mass of solute divided by the total mass of the solution, multiplied by 100. In this case, a 37.0% by mass HCl solution means that there are 37 grams of HCl in every 100 grams of the solution. Understanding this concept is crucial for converting mass to moles in the context of molarity.

Density and Volume Relationship

Density is defined as mass per unit volume, typically expressed in grams per milliliter (g/mL). The density of the hydrochloric acid solution is given as 1.18 g/mL, which allows us to calculate the volume of the solution based on its mass. This relationship is essential for determining how many liters of solution are present when calculating molarity.

Molarity Calculation

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. To find the molarity of the HCl solution, one must first convert the mass of HCl to moles using its molar mass, and then divide by the volume of the solution in liters. This concept is fundamental for solving the problem and determining the correct answer from the provided options.

Recommended video:

Guided course

03:12

Molar Mass Calculation Example

Related Practice

Textbook Question

In which solvent does sodium acetate 1CH COONa2 have the highest solubility (a) Acetone CH3OCH3 (b) Methanol (CH3OH) (c) Ethanol (CH3CH2OH) (d) Pentane (CH3CH2CH2CH2CH3)

views

Textbook Question

Rubbing alcohol is a 90% (by mass) solution of isopropyl alcohol, C3H8O, in water. How many moles of isopropyl alcohol are in 50.0 g of rubbing alcohol? (LO 13.4) (a) 45 mol

views

Textbook Question

Fluoride ion is added to drinking water at low concentra- tions to prevent tooth decay. What mass of sodium fluoride (NaF) should be added to 750 L of water to make a solutionthat is 1.5 ppm in fluoride ion? (LO 13.5) (a) 1.1 g (b) 2.5 g (c) 0.51 g (d) 3.1 g

views

Textbook Question

A 10.0 M aqueous solution of NaOH has a density of 1.33 g/cm3 at 25 C. Calculate the mass percent of the NaOH in the solution (a) 53.2% (b) 30.1% (c) 40.0% (d) 13.3%

Textbook Question

The molarity of a solution of sodium acetate (CH3COONa) at 20 C is 7.5 M. Use the graph showing the solubility of sodium acetate as a function of temperature to describe the solution. The solution is . (LO 13.7, 13.8)

(a) Saturated (b) Unsaturated (c) Supersaturated (d) Colloidal

views

Textbook Question

A scuba diver is breathing compressed air that is 21% (by volume) oxygen at a depth of 66 ft. beneath the surface. The total pressure at this depth is 3.0 atm. The Henry's law constant (k) for O2 in water at a normal body temperature (37 °C ) is 1.93 * 10-3 mol>(L atm). What is the solubilityof O2 in the blood of the diver at 66 ft. beneath the surface?(LO 13.9)(a) 1.22 * 10-3 M (b) 2.76 * 10-2 M(c) 1.22 * 10-1 M (d) 2.76 * 10-1 M