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Chapter 11: Problem 14

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 a 0.580 M CaCl₂ solution containing 1.28 g of solute is \(0.0199\ L\).

Step by step solution

01

Calculate the moles of CaCl₂ needed

To find the moles of CaCl₂, we will use the mass of the solute (1.28 g) and the molar mass of CaCl₂. The molar mass of CaCl₂ is: \(M(CaCl_{2}) = M(Ca) + 2\cdot M(Cl)\) \(M(CaCl_{2}) = 40.08 + 2\cdot 35.45\) \(M(CaCl_{2}) = 110.98 \ g/mol\) Using the mass and the molar mass, we can calculate the moles of CaCl₂: \[\text{moles of CaCl}_{2} = \frac{\text{mass}}{\text{molar mass}}\]
02

Calculate the moles of CaCl₂

Now we can plug in the values and find the moles of CaCl₂: \[\text{moles of CaCl}_{2} = \frac{1.28\ g}{110.98\ g/mol} = 0.01153\ mol\]
03

Use the molarity to find the volume of the solution

We can now use the moles and the given molarity (0.580 M) to find the volume of the solution. We will use the formula: \[M = \frac{n}{V}\] Where M is the molarity, n is the moles of solute, and V is the volume of the solution in liters. We can rearrange this formula to find the volume: \[V = \frac{n}{M}\]
04

Calculate the volume of the 0.580 M CaCl₂ solution

Finally, we can plug in the values of moles and molarity to find the volume: \[V = \frac{0.01153\ mol}{0.580\ M} = 0.01988\ L\] To report the answer with the appropriate number of significant figures, we round the volume to two decimal places (as the molarity has three significant figures and the mass has three significant figures): \[V = 0.0199\ L\] So, the volume of a 0.580 M CaCl₂ solution containing 1.28 g of solute is 0.0199 L.

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Most popular questions from this chapter

From the following: pure water solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water choose the one with the a. highest freezing point. d. lowest boiling point. b. lowest freezing point. e. highest osmotic pressure. c. highest boiling point.

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10 M\) formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of \(0.10 M\) formic acid.

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of \(3.0 \mathrm{M}\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 M\) sodium bicar- bonate.

You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\). Explain what happens and why.

A \(0.500-\mathrm{g}\) sample of a compound is dissolved in enough water to form \(100.0 \mathrm{~mL}\) of solution. This solution has an osmotic pressure of \(2.50\) atm at \(25^{\circ} \mathrm{C}\). If each molecule of the solute disso- ciates into two particles (in this solvent), what is the molar mass of this solute?
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