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Calculate the molarity of a solution that contains (a) \(3.60 \mathrm{~g}\) of sulfuric acid in \(450.0 \mathrm{~mL}\) of solution. (b) \(0.001 \mathrm{~g}\) of iron(II) nitrate in \(12.0 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarity of the sulfuric acid solution is approximately 0.0735 M. The molarity of the iron(II) nitrate solution is approximately 4.63 x 10^-4 M.

Step by step solution

01

Calculate moles of sulfuric acid (H2SO4)

First, find the molar mass of sulfuric acid. H2SO4 has a molar mass of approximately 98.08 g/mol. Next, convert the mass of sulfuric acid to moles by dividing by its molar mass: moles of H2SO4 = 3.60 g / 98.08 g/mol.
02

Convert the volume of the solution to liters

Since molarity is moles per liter, convert the volume of the solution from milliliters to liters: 450.0 mL x (1 L / 1000 mL).
03

Calculate the molarity of the sulfuric acid solution

Molarity (M) is calculated by dividing the number of moles of solute by the volume of solution in liters. Use the values from Step 1 and Step 2.
04

Calculate moles of iron(II) nitrate (Fe(NO3)2)

Find the molar mass of iron(II) nitrate. Fe(NO3)2 has a molar mass of approximately 179.85 g/mol. Convert the mass of iron(II) nitrate to moles by dividing by its molar mass: moles of Fe(NO3)2 = 0.001 g / 179.85 g/mol.
05

Convert the volume of the iron(II) nitrate solution to liters

Convert the volume of the solution from milliliters to liters: 12.0 mL x (1 L / 1000 mL).
06

Calculate the molarity of the iron(II) nitrate solution

Using the values obtained in Step 4 and Step 5, calculate the molarity by dividing the number of moles of solute by the volume of solution in liters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding molar mass is crucial to mastering the concept of molarity. Molar mass, measured in grams per mole (g/mol), represents the weight of one mole of a substance. A mole is basically a count of particles — specifically, Avogadro's number (\(6.022 \times 10^{23}\) particles) of atoms or molecules.

To calculate molar mass, sum the atomic masses of all the atoms in the molecule. For instance, H\textsubscript{2}SO\textsubscript{4} (sulfuric acid) consists of 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms. By using the periodic table, you'd add up the atomic masses of these atoms (approximately 1.01 g/mol for hydrogen, 32.07 g/mol for sulfur, and 16.00 g/mol for oxygen) to find the molar mass. Here's a simple equation to demonstrate: \[ \text{Molar mass of H}_{2}\text{SO}_{4} = (2 \times 1.01) + (1 \times 32.07) + (4 \times 16.00) \text{ g/mol} \]
Moles to Grams Conversion
Once the molar mass is known, conversion between moles and grams becomes straightforward. If you have the mass of a substance in grams and require the amount in moles, divide the mass by the molar mass. This approach gives you the substance's quantity in moles.

Here's the formula you'd use: \[ \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \]
Using the earlier example of sulfuric acid, if you have 3.60 grams (\texttt{g}) of it, and its molar mass is 98.08 g/mol, divide 3.60 g by 98.08 g/mol to find the moles of sulfuric acid.
Solutions Concentration
Molarity, denoted as 'M', measures the concentration of a solution. It's defined as the number of moles of solute present in 1 liter (L) of solution. To calculate the molarity, you'll need the number of moles of the solute and the volume of the solution in liters.

The formula for molarity is simply: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
For example, if you have a sulfuric acid solution containing 3.60 g of acid in a total volume of 450.0 mL, after converting 3.60 g to moles, you would also convert the volume from milliliters to liters (since 1000 mL = 1 L), then apply the molarity formula to determine the solution's concentration in moles per liter.

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Most popular questions from this chapter

Write balanced molecular, ionic, and net ionic equations for the following pairs of reactants. If all ions cancel, indicate that no reaction (N.R.) takes place. (a) sodium sulfite and barium nitrate (b) formic acid \(\left(\mathrm{HCHO}_{2}\right)\) and potassium carbonate (c) ammonium bromide and lead(II) acetate (d) ammonium perchlorate and copper(II) nitrate

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Write balanced ionic and net ionic equations for these reactions. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow\) $$ \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{BaCO}_{3}(s) $$ (b) \(\mathrm{CuCl}_{2}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{NaCl}(a q) $$ (c) \(\mathrm{FeSO}_{4}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Fe}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) (d) \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{NiCl}_{2}(a q) \longrightarrow\) \(\mathrm{AgCl}(s)+\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)\)

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