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Ch.13 - Solutions & Their Properties
Chapter 13, Problem 149f

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (f) What is the identity of the metal M?

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Determine the moles of gas evolved using the ideal gas law: PV = nRT. Convert the given pressure from mm Hg to atm and the volume from mL to L. Use R = 0.0821 L·atm/mol·K and convert the temperature to Kelvin.
Calculate the moles of MClx formed using the freezing point depression formula: ΔT_f = iK_fm, where ΔT_f is the change in freezing point, i is the van't Hoff factor (assume i = 2 for MClx), K_f is the freezing point depression constant for water (1.86 °C·kg/mol), and m is the molality of the solution.
Use the moles of gas evolved to determine the moles of metal M reacted, assuming the gas is hydrogen (H2) and the reaction is M + xHCl → MClx + x/2 H2.
Calculate the molar mass of the metal M by dividing the mass of the metal (1.385 g) by the moles of metal M determined in the previous step.
Identify the metal M by comparing the calculated molar mass to known molar masses of metals.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, and temperature. The ideal gas law (PV=nRT) is particularly relevant here, as it allows us to calculate the number of moles of gas produced during the reaction. Understanding how to manipulate these variables is crucial for determining the amount of metal reacted and the identity of the metal M.
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Combined Gas Law

Molar Mass and Stoichiometry

Molar mass is the mass of one mole of a substance, which is essential for converting between grams and moles. Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. By determining the moles of gas evolved and relating it to the mass of metal M, we can calculate the molar mass of M and identify the metal.
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Freezing Point Depression

Freezing point depression is a colligative property that describes how the addition of a solute lowers the freezing point of a solvent. In this case, the melting point of the solution formed by dissolving MClx in water is given, which can be used to calculate the molality of the solution. This information helps in determining the number of moles of solute, which is necessary for identifying the metal M.
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Related Practice
Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dissolving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (b) What mass of MClx is formed? (a) How many moles of H2 gas are evolved?

Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (c) What is the molality of particles (ions) in the solution of MClx?

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Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (e) What are the formula and molecular weight of MClx?

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Textbook Question

A compound that contains only C and H was burned in excess O2 to give CO2 and H2O. When 0.270 g of the com- pound was burned, the amount of CO2 formed reacted completely with 20.0 mL of 2.00 M NaOH solution according to the equation 2 OH-1aq2 + CO21g2 S CO 2- 1aq2 + H2O1l2 When 0.270 g of the compound was dissolved in 50.0 g of camphor, the resulting solution had a freezing point of 177.9 °C. [#Pure camphor freezes at 179.8 °C and has Kf = 37.7 1°C kg2>mol.] (a) What is the empirical formula of the compound?

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Open Question
Combustion analysis of a 36.72-mg sample of the male hormone testosterone gave 106.43 mg CO2 and 32.10 mg H2O as the only combustion products. When 5.00 mg of testosterone was dissolved in 15.0 mL of a suitable solvent at 25 °C, an osmotic pressure of 21.5 mm Hg was measured. What is the molecular formula of testosterone?
Textbook Question
A solution prepared by dissolving 100.0 g of a mixture of sugar 1C12H22O112 and table salt (NaCl) in 500.0 g of water has a freezing point of - 2.25 °C. What is the mass of each individual solute? Assume that NaCl is completely dissociated.
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