In a mountainous location, the boiling point of pure water is found to be 95°C. How many grams of sodium chloride must be added to 9 kg of water to bring the boiling point back to 101°C? Assume that i = 2.
delta T = i*Kb*m
Substitute and solve for m
m = mols/Kg solvent
Substitute and solve for mols
mol = grams/molar mass. Substitute and solve for grams.
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To find out how many grams of sodium chloride must be added to the water, we need to use the formula for boiling point elevation:
ΔTb = i * m * Kb
Where:
ΔTb is the change in boiling point
i is the Van't Hoff factor (in this case, 2 for sodium chloride)
m is the molality of the solution (moles of solute per kilogram of solvent)
Kb is the molal boiling point constant (a property of the solvent)
First, let's calculate the change in boiling point:
ΔTb = (boiling point of water in pure state) - (boiling point of water in mountainous location)
ΔTb = 100°C - 95°C
ΔTb = 5°C
Next, we need to find the molality of the solution. Molality (m) is defined as moles of solute per kilogram of solvent. In this case, the solute is sodium chloride (NaCl) and the solvent is water.
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
To calculate the molality, we need to convert the given mass of water to moles:
Number of moles of water = mass of water / molar mass of water
Number of moles of water = 9000 g / 18 g/mol
Number of moles of water = 500 mol
Now we can calculate the molality:
m = moles of solute / kg of solvent
m = (mass of NaCl / molar mass of NaCl) / (mass of water / 1000)
m = (mass of NaCl / 58.5) / 9
Finally, we can rearrange the boiling point elevation equation to solve for the mass of NaCl:
mass of NaCl = (ΔTb * m * M) / i
mass of NaCl = (5 * ((mass of NaCl / 58.5) / 9) * 58.5) / 2
Now we can solve for the mass of NaCl. Let's rearrange the equation to isolate the mass of NaCl:
mass of NaCl = (5 * ((mass of NaCl / 58.5) / 9) * 58.5) / 2
Multiplying both sides by 2:
2 * mass of NaCl = 5 * ((mass of NaCl / 58.5) / 9) * 58.5
Multiplying both sides by 9:
18 * mass of NaCl = 5 * (mass of NaCl / 58.5) * 58.5
Dividing both sides by 5 * 58.5:
3.06 * mass of NaCl = mass of NaCl / 58.5
Dividing both sides by mass of NaCl:
3.06 = 1 / 58.5
Dividing both sides by 3.06:
1 = 58.5 / 3.06
mass of NaCl = 58.5 / 3.06
Therefore, the mass of sodium chloride that must be added to 9 kg of water to bring the boiling point back to 101°C is approximately 19.12 grams.