Dosage Specification: What Volume of 0.20mnacl(aq) Contains 10.0g of NACL (molar mass 58g/mol)?

What Volume of 0.20mnacl(aq) Contains 10.0g of NACL (molar mass 58g/mol)?

Have you ever wondered how to calculate the volume of a solution containing a specific amount of solute? In this article, I’ll walk you through the steps to determine the volume of a 0.20M NaCl(aq) solution that contains 10.0g of NaCl (molar mass 58g/mol). By following these calculations, you’ll be able to accurately measure and prepare solutions for various chemical experiments.

To begin, we need to convert the mass of NaCl into moles. Since we know the molar mass is 58g/mol, dividing 10.0g by 58g/mol gives us approximately 0.172 mol of NaCl. Next, we can use the equation Molarity = Moles/Volume (in liters) to find the volume. Since we’re given that the solution has a concentration of 0.20M (moles per liter), we can rearrange the equation to solve for volume: Volume = Moles/Molarity. Plugging in our values, Volume = 0.172 mol / 0.20M, which gives us an answer of approximately 0.86L. By applying these calculations, you can determine the required volume of a specific solute in a given concentration solution accurately and efficiently. Understanding these concepts will prove invaluable in conducting precise chemical experiments and measurements. Stay tuned as I provide further insights into chemistry topics and guide you through other intriguing calculations!

Calculating Molar Mass of NaCl

Let’s dive into the fascinating world of chemistry and explore how to calculate the molar mass of NaCl. The molar mass, also known as molecular weight, is a crucial concept in understanding chemical reactions and stoichiometry. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). In this section, we’ll break down the steps to determine the molar mass of NaCl. To calculate the molar mass of NaCl, we first need to know the atomic masses of its constituent elements: sodium (Na) and chlorine (Cl). Sodium has an atomic mass of approximately 23 g/mol, while chlorine has an atomic mass close to 35.5 g/mol. Next, we multiply these atomic masses by their respective subscripts in the formula for NaCl. Since there is only one atom each of sodium and chlorine in NaCl, we simply add their individual masses together to obtain the total molar mass: Molar Mass(NaCl) = Atomic Mass(Na) + Atomic Mass(Cl) = 23 g/mol + 35.5 g/mol ≈ 58.5 g/mol Therefore, the molar mass of NaCl is approximately 58.5 g/mol. Now that we have calculated the molar mass of NaCl, let’s apply this knowledge to solve a problem: determining what volume of a 0.20 M solution contains 10.0 grams of NaCl. To find the volume, we can use the formula: Volume = (mass / molar mass) * concentration Substituting in our values: Volume = (10.0 g / 58.5 g/mol) * 0.20 mol/L After performing the calculation using appropriate units for grams and liters, we will arrive at our desired answer. Remember that understanding molar mass is essential for various aspects of chemistry, such as stoichiometry and balancing chemical equations. It allows us to relate the amount of a substance in moles to its mass in grams or volume in liters.

Determining the Number of Moles in 10.0g of NaCl

To determine the number of moles in 10.0g of NaCl, we need to convert the given mass into moles using the molar mass of NaCl. The molar mass of NaCl is 58g/mol. First, let’s calculate how many moles are present in 1 gram of NaCl. Since 1 mole of any substance contains Avogadro’s number (6.022 x 10^23) of particles, we can use this conversion factor to find the number of moles. 1 mol NaCl = 58 g Dividing both sides by 58 gives us: 1 g NaCl = 1/58 mol Now that we know that there is approximately 1/58 mol in every gram of NaCl, we can calculate the number of moles in our given sample: Number of moles = Mass / Molar mass Number of moles = 10.0g / 58g/mol Number of moles ≈ 0.1724 mol Therefore, there are approximately 0.1724 moles in a sample containing 10.0g of NaCl. This calculation allows us to understand the amount (in terms of moles) present in a given mass quantity and is an essential step when working with chemical reactions or stoichiometry problems.