The molar concentration will be greater than 0.01 M .
Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.
The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ = i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration =
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = i
m we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ = 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Learn more about freezing point here;
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<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
To calculate the heat absorbed by the water, we use the equation:
where,
q = heat absorbed
m = mass of water = 1500 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature =
Putting values in above equation, we get:
Hence, the amount of heat required to warm given amount of water is 470.9 kJ