Problem 67 - Chemistry The Central Science

Question

Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},\) (b) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},\) (c) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

Step-by-Step Solution

Answer

(a) 5.21 g of KBr (b) 0.0654 M Ca(NO\(_3\))\(_2\) (c) 10.2 mL of 1.50 M Na\(_3\)PO\(_4\)

Step 1: Calculate moles of solute in KBr solution

To determine the moles of solute in the KBr solution, use the formula: moles of solute = molarity × volume moles of \(KBr = 0.175\,M × 0.250\,L = 0.04375\,mol\)

Step 2: Calculate grams of solute in KBr solution

Now, we'll convert the moles of solute to grams using the molar mass of KBr. Molar mass of KBr = 39.1 g/mol (for K) + 79.9 g/mol (for Br) = 119 g/mol Grams of KBr = moles × molar mass = \(0.04375\,mol × 119\,g/mol = 5.21\,g\) So, there are 5.21 grams of KBr solute.

Step 3: Calculate moles of solute in Ca(NO\(_3\))\(_2\) solution

First, find the moles of solute using the mass and molar mass of Ca(NO\(_3\))\(_2\). Molar mass of Ca(NO\(_3\))\(_2\) = 40.08 g/mol (for Ca) + 2 × (14.01 g/mol (for N) + 3 × 16.00 g/mol (for O)) = 164.10 g/mol Moles of Ca(NO\(_3\))\(_2\) = mass / molar mass = \(14.75\,g / 164.10\,g/mol = 0.0899\,mol\)

Step 4: Calculate the molar concentration of the Ca(NO\(_3\))\(_2\) solution

Now, we'll calculate the molar concentration using the formula: molarity = moles of solute / volume Molarity of Ca(NO\(_3\))\(_2\) = \(0.0899\,mol / 1.375\,L = 0.0654\,M\) So, the molar concentration of the Ca(NO\(_3\))\(_2\) solution is 0.0654 M.

Step 5: Calculate moles of solute in Na\(_3\)PO\(_4\) solution

First, find the moles of solute using the mass and molar mass of Na\(_3\)PO\(_4\). Molar mass of Na\(_3\)PO\(_4\) = 3 × 22.99 g/mol (for Na) + 30.97 g/mol (for P) + 4 × 16.00 g/mol (for O) = 163.94 g/mol Moles of Na\(_3\)PO\(_4\) = mass / molar mass = \(2.50\,g / 163.94\,g/mol = 0.0153\,mol\)

Step 6: Calculate the volume of Na\(_3\)PO\(_4\) solution

Now, find the volume in liters using the formula: volume = moles of solute / molarity Volume of Na\(_3\)PO\(_4\) = \(0.0153\,mol / 1.50\,M = 0.0102\,L\)

Step 7: Convert the volume to milliliters

Finally, convert the volume from liters to milliliters. Volume of Na\(_3\)PO\(_4\) = \(0.0102\,L × 1000\,mL/L = 10.2\,mL\) So, the volume of 1.50 M Na\(_3\)PO\(_4\) in milliliters that contains 2.50 g of solute is 10.2 mL.

Key Concepts

Molarity

Molarity is a measure of concentration. It tells us how much solute is present in a certain volume of solution. Molarity is expressed in terms of moles of solute per liter of solution, with the unit being mol/L or simply M. For example, if a solution has a molarity of 1 M, it means there is 1 mole of solute dissolved in 1 liter of solution. You calculate molarity by dividing the number of moles of solute by the volume of the solution in liters. This formula is typically written as: \[\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]To better understand, consider a solution containing 0.0899 mol of calcium nitrate (Ca(NO\(_3\))\(_2\)) in 1.375 liters of solution. Using our formula, the molarity is calculated as 0.0654 M. Molarity helps chemists specify concentrations precisely, enabling accurate predictions of how a solution will react with other substances in further chemical processes.

Moles of Solute

The mole is a fundamental unit in chemistry that measures the amount of substance. One mole is defined as 6.022 × 10\(^23\) particles of the substance, be those atoms, molecules, or ions. The concept of moles allows chemists to count particles by weighing them, facilitating laboratory calculations and chemical reactions. To find the moles of solute in a solution, you can use the mass of the solute and its molar mass. For example, for 14.75 grams of calcium nitrate Ca(NO\(_3\))\(_2\), where the molar mass is 164.10 g/mol, the moles are calculated as follows:\[\text{Moles of } Ca(NO\(_3\))\(_2\) = \frac{14.75\, \text{g}}{164.10\, \text{g/mol}} = 0.0899\, \text{mol}\]This conversion is crucial as it allows you to link the measurable mass of a substance to the scaling factor needed to apply it in balanced equations for reactions. It's the bridge connecting mass to moles and then to molarity.

Molar Mass Calculation

The molar mass of a compound is the mass of one mole of that compound, usually expressed in grams per mole (g/mol). Molar mass is calculated by adding up the atomic masses of all the atoms in a chemical formula. When calculating the molar mass of a compound like KBr, you add the atomic masses of potassium (K) and bromine (Br):\[\text{Molar mass of } KBr = 39.1\, \text{g/mol (for K)} + 79.9\, \text{g/mol (for Br)} = 119\, \text{g/mol}\]This gives you a concrete number that links chemical formulas directly to measurable masses. In practice, if you need the mass of a certain number of moles of a compound, you multiply the number of moles by this molar mass. For example, to find the mass of 0.04375 moles of KBr, you calculate:\[\text{Grams of } KBr = 0.04375\, \text{mol} \times 119\, \text{g/mol} = 5.21\, \text{g}\]Using the molar mass, you can convert moles back into grams, which is essential for practical laboratory work.