How many grams Kl are needed to prepare 100.mL of 0.55M solution?
1 answer:
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Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion +
Where, negative sign signifies heat loss
Or,
Heat of fusion +
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>