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Chapter 5: Problem 48

Calculate the density of hydrogen bromide (HBr) gas in grams per liter at \(733 \mathrm{mmHg}\) and \(46^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The density of HBr gas under the given conditions is calculated to be approximately 2.40 g/L.

Step by step solution

01

Determine the Molar Mass of Hydrogen Bromide

Hydrogen bromide is made up of one atom of hydrogen (H) and one atom of bromine (Br). The molar mass of hydrogen is approximately 1 g/mol, and that of bromine is approximately 79.9 g/mol. Apply these values to calculate the molar mass of HBr. \( M_{HBr} = 1g/mol + 79.9g/mol = 80.9 g/mol \)
02

Convert Temperature from Celsius to Kelvin

The ideal gas law requires temperature to be in Kelvin. You can convert from Celsius to Kelvin using the formula \( K = C + 273.15 \), where \( K \) is temperature in Kelvin and \( C \) is temperature in Celsius. Therefore, the temperature in Kelvin is \( 46掳C + 273.15 = 319.15 K \).
03

Convert Pressure from mmHg to atm

We also need to convert the pressure from millimeters of mercury (mmHg) to atmospheres (atm) because the ideal gas law commonly uses options for the pressure in atm. To convert mmHg to atm, use the conversion rate 1 atm = 760 mmHg. Therefore, the pressure in atm is \( 733 mmHg / 760 mmHg/atm = 0.964 atm \).
04

Calculate the Density of HBr Gas

The general formula for gas density is \( d = \frac{MP}{RT} \), where \( d \) is the density, \( M \) is the molar mass, \( P \) is the pressure, \( R \) is the ideal gas constant (which equals to 0.0821 L鈭檃tm/mol鈭橩), and \( T \) is the temperature. Substituting the known values into the formula will yield \( d = \frac{(80.9 g/mol) 脳 (0.964 atm)}{(0.0821 L鈭檃tm/mol鈭橩) 脳 (319.15 K)} \).

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Most popular questions from this chapter

Dry ice is solid carbon dioxide. A 0.050 -g sample of dry ice is placed in an evacuated \(4.6-\mathrm{L}\) vessel at \(30^{\circ} \mathrm{C}\). Calculate the pressure inside the vessel after all the dry ice has been converted to \(\mathrm{CO}_{2}\) gas.

A sample of air occupies \(3.8 \mathrm{~L}\) when the pressure is \(1.2 \mathrm{~atm} .\) (a) What volume does it occupy at \(6.6 \mathrm{~atm} ?\) (b) What pressure is required in order to compress it to \(0.075 \mathrm{~L} ?\) (The temperature is kept constant.)

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