Question

Upon heating a litre of semi-molar \(\mathrm{HCl}\) solution, \(2.675 \mathrm{~g}\) of hydrogen chloride is lost and the volume of the solution shrinks to \(750 \mathrm{ml}\). The molarity of resultant solution is (a) \(0.569 \mathrm{M}\) (b) \(0.5 \mathrm{M}\) (c) \(0.42 \mathrm{M}\) (d) \(1.707 \mathrm{M}\)

Short answer

Answer: The molarity of the resulting solution after heating is 0.569 M.

Step-by-step solution

Calculate the moles of HCl in the initial solution

First, we need to find out how many moles of HCl were in the initial solution. We are given that the initial solution was semi-molar, which means its molarity was 0.5M. The moles of HCl in the initial solution can be found using the molarity equation mentioned above: \(M = \dfrac{n}{V}\) Rearranging to find n: \(n = M \times V\) Here, M = 0.5 M V = 1 L (given) \(n_{initial} = 0.5 M \times 1 L = 0.5\) moles So, the initial solution had 0.5 moles of HCl.

Calculate the mass loss in moles

We are given that 2.675 g of hydrogen chloride was lost when heated. To find the moles of HCl lost, we need to convert this mass into moles by dividing the mass by the molar mass of HCl. Molar mass of HCl = 36.5 g/mol. \(n_{lost} = \dfrac{Mass_{lost}}{Molar_{mass}}\) So, \(n_{lost} = \dfrac{2.675 g}{36.5 g/mol} = 0.07329\) moles

Calculate the moles of HCl left

Now, subtract the moles of HCl lost from the initial moles of HCl to find the moles of HCl left after heating: \(n_{final} = n_{initial} - n_{lost}\) So, \(n_{final} = 0.5 - 0.07329 = 0.42671\) moles

Calculate the final molarity of the solution

Finally, we can calculate the molarity of the final solution using the moles of HCl left and the final volume of the solution (750 ml = 0.75 L): \(M_{final} = \dfrac{n_{final}}{V_{final}}\) So, \(M_{final} = \dfrac{0.42671 \, moles}{0.75 L} = 0.569 M\) Therefore, the molarity of the resulting solution after heating is 0.569 M, which corresponds to answer choice (a).

Key concepts

Understanding Hydrochloric Acid (HCl)

Hydrochloric acid, often abbreviated as HCl, is a straightforward yet powerful compound. It consists of a hydrogen (H) atom bonded to a chlorine (Cl) atom. This simple diatomic molecule makes HCl a classic example of a strong acid.

Strong acids fully disassociate in water, releasing hydrogen ions which contribute to the acidity of the solution. This characteristic is crucial in various fields, from industrial processes to biological systems, influencing reactions and conditions profoundly.

HCl is a common laboratory acid, known for its use in titration, pH regulation, and chemical synthesis. In our daily lives, it is found in the stomach, aiding digestion by creating an acidic environment. HCl plays multiple roles due to its straightforward chemistry and pronounced reactivity.

Exploring Moles and Molar Mass

In chemistry, understanding moles and molar mass is fundamental.The concept of a mole allows chemists to count the number of particles, like atoms or molecules, using a consistent unit.One mole equals Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.

The molar mass of a substance is the mass of one mole of that substance, measured in grams per mole (g/mol).For hydrochloric acid (HCl), the molar mass is calculated by adding the atomic masses of hydrogen (1 g/mol) and chlorine (35.5 g/mol), totaling 36.5 g/mol.

Knowing how to convert between mass and moles is essential for accurate calculations in chemistry.This ability enables chemists to predict the outcomes of reactions and determine the necessary quantities of reactants and products.

Calculating Solution Concentration

Solution concentration, often expressed as molarity (M), tells us how much solute is dissolved in a solution. Molarity is defined as the number of moles of solute divided by the volume of the solution in liters.

For example, a semi-molar solution of HCl has a molarity of 0.5 M, meaning 0.5 moles of HCl is dissolved per liter of solution. Changes in concentration can result from either adding more solute or altering the solution's volume.

In the context of heating a solution, water might evaporate, concentrating the remaining solute as the volume decreases. This process needs precise calculations to ensure desired concentrations are maintained for specific applications. Using molarity is crucial to prepare solutions in laboratories or industrial settings, ensuring reactions proceed as intended.