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Chapter 8: Problem 7

At \(373 \mathrm{~K}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} / \mathrm{m} 1\) and that of water vapour is \(0.0006 \mathrm{~g} / \mathrm{ml}\), then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) \(6 \mathrm{ml}\) (b) \(60 \mathrm{ml}\) (c) \(0.6 \mathrm{ml}\) (d) \(0.06 \mathrm{ml}\)

Short Answer

Expert verified
Answer: The volume occupied by water molecules in 1 litre of steam at 373K and 1 atm is 0.6 ml.

Step by step solution

01

Find the mass of 1000 ml of steam

To find the mass of 1000 ml of steam, we will use the given density of water vapour, which is 0.0006 g/ml. Multiply it by 1000 ml to obtain the mass: Mass = Density * Volume Mass = (0.0006 g/ml) * (1000 ml) Mass = 0.6 g
02

Calculate the number of moles of water in steam

To find the number of moles, we will use the molar mass of water which is 18.01528 g/mol. We will then divide the mass of steam by the molar mass: Number of moles = Mass / Molar mass Number of moles = 0.6 g / 18.01528 g/mol Number of moles ≈ 0.0333 mol
03

Calculate the mass of water molecules in steam

Now, let's find the mass of water molecules only in steam. We'll do this by multiplying the number of moles by the molar mass of water: Mass of water molecules = Number of moles * Molar mass Mass of water molecules = 0.0333 mol * 18.01528 g/mol Mass of water molecules ≈ 0.6 g
04

Calculate the volume occupied by water molecules in steam

Now, we can find the volume occupied by the water molecules in steam using the density of liquid water, which is 1.0 g/ml: Volume = Mass / Density Volume = 0.6 g / 1.0 g/ml Volume = 0.6 ml Therefore, the volume occupied by water molecules in 1 litre of steam at 373K and 1 atm is 0.6 ml (Option c).

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Most popular questions from this chapter

Air contains \(20 \% \mathrm{O}_{2}\), by volume. What volume of air is needed at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) for complete combustion of \(80 \mathrm{~g}\) methane? (a) 101 (b) 501 (c) 2241 (d) 11201

One mole of a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) requires exactly \(20 \mathrm{~g}\) of \(\mathrm{NaOH}\) in solution for complete conversion of all the \(\mathrm{CO}_{2}\) into \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) How many grams more of \(\mathrm{NaOH}\) would it require for conversion into \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) if the mixture (one mole) is completely oxidized to \(\mathrm{CO}_{2} ?\) (a) \(60 \mathrm{~g}\) (b) \(80 \mathrm{~g}\) (c) \(40 \mathrm{~g}\) (d) \(20 \mathrm{~g}\)

A volume of \(200 \mathrm{ml}\) of oxygen is added to \(100 \mathrm{ml}\) of a mixture containing \(\mathrm{CS}_{2}\) vapour and \(\mathrm{CO}\), and the total mixture is burnt. After combustion, the volume of the entire mixture is \(245 \mathrm{ml}\). Calculate the volume of the oxygen that remains (a) \(67.5 \mathrm{ml}\) (b) \(125.0 \mathrm{ml}\) (c) \(200.0 \mathrm{ml}\) (d) \(100.0 \mathrm{ml}\)

What mass of carbon disulphide, \(\mathrm{CS}_{2}\), can be completely oxidized to \(\mathrm{SO}_{2}\) and \(\mathrm{CO}_{2}\) by the oxygen liberated when \(325 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\) react with water? (a) \(316.67 \mathrm{~g}\) (b) \(52.78 \mathrm{~g}\) (c) \(633.33 \mathrm{~g}\) (d) \(211.11 \mathrm{~g}\)

A \(1.50 \mathrm{~g}\) sample of potassium bicarbonate having \(80 \%\) purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is (a) \(3.72 \mathrm{~g}\) (b) \(0.72 \mathrm{~g}\) (c) \(0.372 \mathrm{~g}\) (d) \(0.186 \mathrm{~g}\)
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