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bezimeni [28]
3 years ago
8

The density of water is 1.00 g/cm3. The density of ice is 0.92 g/cm3. By what percent is the volume increased when water is froz

en at 0°C? What is the final volume if 500 mL of water is completely frozen?
Chemistry
2 answers:
cestrela7 [59]3 years ago
8 0

Answer:

At the same temperature of 0°C, the density (mass per volume) of ice is 0.9187 gram per cubic centimeter (g cm-3 or g/cm3) while that of liquid water is 0.9998 g cm-3 (Cohen et al. 2003). The lesser density means that ice contains lesser mass (quantity of matter) per unit of volume. It also means that the molecules of water are less compressed per unit volume of ice. Another way of saying it is that ice is less compact than liquid water having the same volume.

4vir4ik [10]3 years ago
7 0

Answer:

Volume increase by 8.70% and volume of 500 mL of water when completely frozen is 543 mL

Explanation:

  • Density is the ratio of mass to volume
  • Let's say there are m g of water. So volume of m g of water is equal to \frac{m}{1.00}cm^{3} or m cm^{3}
  • If m g of water is completely frozen to ice then volume of m g of ice is equal to \frac{m}{0.92}cm^{3}
  • So, percentage of increases in volume = \frac{(\frac{m}{0.92})-(m)}{m}\times 100 % = 8.70 %
  • Mass of 500 mL of water = (1.00g/cm^{3})\times 500cm^{3}=500g (1mL=1cm^{3})
  • So volume of 500 mL of water when it is completely frozen = \frac{500g}{0.92g/cm^{3}}=543cm^{3}=543 mL
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P = Pressure of the gaseous mixture = 1.00 atm

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n = number of moles of the gaseous mixture = ?

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Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

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Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

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Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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