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Chapter 11: Problem 16

What volume of a \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) contains 1.28 \(\mathrm{g}\) solute?

Short Answer

Expert verified
The volume of the \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) that contains 1.28 g solute is approximately 19.88 mL.

Step by step solution

01

Calculate the molar mass of CaCl₂

First, we need to find the molar mass of CaCl₂. The molar masses of \(\mathrm{Ca}\), \(\mathrm{Cl}\) are 40.08 g/mol and 35.45 g/mol respectively. So, we can find the molar mass of CaCl₂ as follows: Molar mass of CaCl₂ = (1 × 40.08) + (2 × 35.45) = 40.08 + 70.90 = 110.98 g/mol.
02

Convert the mass of solute from grams to moles

Now, let's convert the mass of solute (1.28 g) to moles using the molar mass of CaCl₂ (110.98 g/mol). Number of moles = (mass of solute) / (molar mass of CaCl₂) = 1.28 g / 110.98 g/mol = 0.01153 moles
03

Use the definition of molarity to find the volume of the solution

The definition of molarity (M) is: M = moles of solute / volume of solution (L) We have the molarity of the solution (0.580 M) and the moles of solute (0.01153 moles). Now, we can solve for the volume of the solution (in liters): Volume of solution (L) = moles of solute / molarity = 0.01153 moles / 0.580 M = 0.019876 L
04

Convert the volume to milliliters

Lastly, we will convert the volume from liters to milliliters for easier reading: Volume of solution (mL) = 0.019876 L × 1000 mL/L = 19.88 mL So, the volume of the 0.580-M solution of CaCl₂ that contains 1.28 g of solute is approximately 19.88 mL.

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Most popular questions from this chapter

The vapor pressure of pure benzene is 750.0 torr and the vapor pressure of toluene is 300.0 torr at a certain temperature. You make a solution by pouring “some” benzene with “some” toluene. You then place this solution in a closed container and wait for the vapor to come into equilibrium with the solution. Next, you condense the vapor. You put this liquid (the condensed vapor) in a closed container and wait for the vapor to come into equilibrium with the solution. You then condense this vapor and find the mole fraction of benzene in this vapor to be 0.714. Determine the mole fraction of benzene in the original solution assuming the solution behaves ideally.

Explain the terms isotonic solution, crenation, and hemolysis.

An aqueous solution is 1.00\(\% \mathrm{NaCl}\) by mass and has a density of 1.071 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) . The observed osmotic pressure of this solution is 7.83 atm at \(25^{\circ} \mathrm{C}\) . a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution.

Erythrocytes are red blood cells containing hemoglobin. In a saline solution they shrivel when the salt concentration is high and swell when the salt concentration is low. In a \(25^{\circ} \mathrm{C}\) aqueous solution of NaCl, whose freezing point is \(-0.406^{\circ} \mathrm{C},\) erythrocytes neither swell nor shrink. If we want to calculate the osmotic pressure of the solution inside the erythrocytes under these conditions, what do we need to assume? Why? Estimate how good (or poor) of an assumption this is. Make this assumption and calculate the osmotic pressure of the solution inside the erythrocytes.

The vapor pressure of a solution containing 53.6 \(\mathrm{g}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in 133.7 \(\mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) . Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.
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