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Chapter 11: Problem 126

A solid mixture contains \(\mathrm{MgCl}_{2}\) and NaCl. When \(0.5000 \mathrm{~g}\) of this solid is dissolved in enough water to form \(1.000 \mathrm{~L}\) of solution, the osmotic pressure at \(25.0^{\circ} \mathrm{C}\) is observed to be \(0.3950\) atm. What is the mass percent of \(\mathrm{MgCl}_{2}\) in the solid? (Assume ideal behavior for the solution.)

Short Answer

Expert verified
The mass percent of MgCl₂ in the solid mixture is approximately \(38.68\%\).

Step by step solution

01

Write down the formula for osmotic pressure.

We will be using the osmotic pressure formula to relate the moles of solute particles to the osmotic pressure: Osmotic pressure (π) = (n/V) × R × T where: n = moles of solute particles V = volume of the solution R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature (in Kelvin)
02

Express the moles of solute particles in terms of concentration of MgCl₂ and NaCl.

Let x be the mass of MgCl₂ in the 0.5000 g mixture. Then, the mass of NaCl is (0.5000 - x) g. We can express the moles of solute particles (n) in terms of concentration (C) of MgCl₂ and NaCl as follows: n = (x / M₁) × 2 + ((0.5000 - x) / M₂) where: M₁ = molar mass of MgCl₂ (95.211 g/mol) M₂ = molar mass of NaCl (58.443 g/mol) 2 is a factor considering that MgCl₂ and NaCl dissociate into 2 and 1 particles in the solution, respectively.
03

Convert the temperature to Kelvin and substitute values into the osmotic pressure formula.

To convert the given temperature from Celsius to Kelvin, we add 273.15: T = 25.0°C + 273.15 = 298.15 K Now, substitute the values into the osmotic pressure formula: 0.3950 atm = [(x / 95.211) × 2 + (0.5000 - x) / 58.443] × (0.0821 L·atm/mol·K) × 298.15 K / 1.000 L
04

Solve for x.

To solve for x, we can now rearrange the equation and solve for x using algebra: x = (0.3950 atm × 1.000 L) / [(0.0821 L·atm/mol·K) × 298.15 K] - (0.5000 - x) / 58.443 × 95.211 / 2 Solve for x: x ≈ 0.1934 g
05

Calculate the mass percent of MgCl₂.

Now, we can calculate the mass percent of MgCl₂ in the solid mixture as follows: Mass percent of MgCl₂ = (Mass of MgCl₂ / Total mass) × 100% Mass percent of MgCl₂ = (0.1934 g / 0.5000 g) × 100% Mass percent of MgCl₂ ≈ 38.68% The mass percent of MgCl₂ in the solid mixture is approximately 38.68%.

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Most popular questions from this chapter

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C}\). You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C}\) ) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at \(25^{\circ} \mathrm{C}\) ) that has a boiling point of \(102.0^{\circ} \mathrm{C}\). What will happen to the plant cells in the leaf?

Calculate the freezing point and boiling point of an antifreeze solution that is \(50.0 \%\) by mass of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) in water. Ethylene glycol is a nonelectrolyte.

A \(0.15-\mathrm{g}\) sample of a purified protein is dissolved in water to give \(2.0 \mathrm{~mL}\) of solution. The osmotic pressure is found to be \(18.6\) torr at \(25^{\circ} \mathrm{C}\). Calculate the protein's molar mass.

Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation.) a. \(0.050 \mathrm{~m} \mathrm{MgCl}_{2}\) b. \(0.050 \mathrm{~m} \mathrm{FeCl}_{3}\)

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner \(\mathrm{R} . \mathrm{B}\). Woodward. It is used as a tranquilizer and sedative. When \(1.00 \mathrm{~g}\) reserpine is dissolved in \(25.0 \mathrm{~g}\) camphor, the freezing-point depression is \(2.63^{\circ} \mathrm{C}\left(K_{\mathrm{f}}\right.\) for camphor is \(40 .{ }^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) ). Calculate the molality of the solution and the molar mass of reserpine.
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