/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 121 A chloride of a metal (M) has \(... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 121

A chloride of a metal (M) has \(65.5 \%\) of chlorine. 100 \(\mathrm{ml}\) of vapour of the chloride of metal at STP weighs \(0.72 \mathrm{~g}\). The molecular formula of this metal chloride is (a) \(\mathrm{MCl}_{3}\) (b) \(\mathrm{MCl}_{4}\) (c) \(\mathrm{M}_{2} \mathrm{Cl}_{3}\) (d) \(\mathrm{MCl}_{3}\)

Short Answer

Expert verified
The molecular formula of the metal chloride is \(MCl_3\).

Step by step solution

01

Determine Molar Mass of Chloride

From the given information, the chloride of metal contains 65.5% chlorine by mass. This implies that in 100g of the compound, 65.5g is chlorine, and 34.5g is metal M.
02

Calculate Molar Mass of the Compound

The mass of 100 ml vapor at STP can be used to determine the molar mass. At STP, 1 mol of gas is 22.4 L or 22400 ml. Using direct proportionality:\[\frac{0.72 \, ext{g}}{100 \, ext{ml}} = \frac{x \, ext{g}}{22400 \, ext{ml}} \]This results in:\[x = \frac{0.72 \, ext{g} \times 22400 \, ext{ml}}{100 \, ext{ml}} = 161.28 \, ext{g/mol}\]Hence, the molar mass of the chloride is 161.28 g/mol.
03

Calculate Molar Mass of Chlorine in Compound

The mass that is chlorine is 65.5% of 161.28 g/mol. Therefore:\[\text{mass of chlorine} = 0.655 \times 161.28 \, ext{g/mol} = 105.56 \, ext{g/mol}\]Now, calculate the amount of chlorine in moles:\[\frac{105.56}{35.5} \approx 3 \]Chlorine exists as \(Cl_3\) based on approximately 3 moles.
04

Determine Molar Mass of Metal M

Subtract the molar mass of chlorine from the total molar mass to find the molar mass of the metal M:\[\text{molar mass of M} = 161.28 \, ext{g/mol} - 105.56 \, ext{g/mol} = 55.72 \, ext{g/mol}\]Thus, the molar mass of metal M is approximately 55.72 g/mol.
05

Identify the Molecular Formula

We have \(MCl_3\) from our calculations since there are 3 moles of chlorine. Therefore, the molecular formula for the chloride of metal is \(MCl_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Molar mass is a fundamental concept in chemistry used to determine how much one mole of a substance weighs. It is typically expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you need to know the atomic masses of the elements that make up the compound and how many atoms of each element are present. When approaching a problem involving molar mass, follow these steps:
  • Identify all the elements in the compound.
  • Find the atomic mass of each element, usually available from the periodic table.
  • Multiply the atomic mass of each element by the number of times the element appears in the compound.
  • Add up all the masses to get the total molar mass of the compound.
In the exercise, given that the vapor weighs 0.72 g/100 ml, we used it to find the molar mass by understanding that 1 mole of gas at standard temperature and pressure (STP) occupies 22.4 liters. This involves converting the measure from ml to liters and applying direct proportionality to find the molar mass as 161.28 g/mol.
Vapor Density
Vapor density is a useful parameter in determining the molar mass of gaseous substances. It is calculated as the mass of a certain volume of vapor compared to the same volume of hydrogen gas under the same conditions. However, it can also relate to how much a particular volume of vapor weighs in relation to standard conditions, usually at STP. Understanding vapor density helps chemists infer the molar mass from a known vapor weight per unit volume. At STP, 1 mole of any gas occupies 22.4 liters. This standard allows the scientists to deduce relationships between the volume of gas measured and the molar mass. In our problem, 100 ml of gas weighed 0.72 g. By scaling up this relationship to a full mole's volume (22,400 ml for gases at STP), we deduced the molar mass of the chloride compound. Essentially, multiplying the vapor density by the equation for STP volume allowed us to compute the mass of the entire mole of the chloride compound accurately.
Chlorine Percentage Composition
Percentage composition specifies how much of each element is in a compound by mass. For chlorine in this problem, this means calculating the fraction of the compound's mass that chlorine constitutes, converting it to a percentage. This is calculated by:
  • Taking the mass of the element in 100g of the compound.
  • Dividing it by the total molar mass of the compound.
  • Multiplying by 100 to express it as a percentage.
In this exercise, the given assumption is that 65.5% of the compound is chlorine. Meaning, in 100 g of the chloride, chlorine accounts for 65.5 g. This information was used as a basis to further deduce the molar mass distribution of chlorine within the compound and ultimately identify the number of chlorine atoms per molecule as three, leading to the formula \(MCl_3\). Calculating percentage composition is crucial, as it helps verify that the alloy or compound ratios align with known or expected configurations based on empirical and theoretical data.

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Most popular questions from this chapter

The equivalent weight of \(\mathrm{MnSO}_{4}\) is half its molecular weight when it is converted to (a) \(\mathrm{MnO}\) (b) \(\mathrm{MnO}_{4}^{2-}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{MnO}_{4}^{-}\)

The hydrated salt \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O}\) undergoes \(63 \%\) loss in mass on heating and becomes anhydrous. The value of \(\mathrm{n}\) is (a) 4 (b) 6 (c) 8 (d) 10

Sodium bicarbonate on heating decomposes to form sodium carbonate, \(\mathrm{CO}_{2}\) and water. If \(0.2\) moles of sodium bicarbonate is completely decomposed, how many moles of sodium carbonate is formed? (a) \(0.1\) (b) \(0.2\) (c) \(0.05\) (d) \(0.025\)

From the complete decomposition of \(20 \mathrm{~g} \mathrm{CaCO}_{3}\) at STP the volume of \(\mathrm{CO}_{2}\) obtained is (a) \(2.24 \mathrm{~L}\) (b) \(4.48 \mathrm{~L}\) (c) \(44.8 \mathrm{~L}\) (d) \(48.4 \mathrm{~L}\)

The weight of a single atom of oxygen is (a) \(5.057 \times 10^{23} \mathrm{~g}\) (b) \(1.556 \times 10^{23} \mathrm{~g}\) (c) \(2.656 \times 10^{-23} \mathrm{~g}\) (d) \(4.538 \times 10^{-23} \mathrm{~g}\)
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