Assuming complete dissociation of the solute, how many grams of KNO_3 must be added to 275mL of water to produce a solution that freezes at -14.5C? The freezing point for pure water is 0.0 C and K_f is equal to 1.86 C.

Delta T= Kf*m*i

i=2
m=3.90
solvent 275=.275L

Moles=1.07

Can anyone help plz? Stuck

moles = grams/molar mass

You have moles and molar mass. Calculate grams.

To solve this problem, you first need to calculate the moles of KNO3 required to produce the desired freezing point depression.

1. Calculate delta T (change in freezing point):
ΔT = -14.5°C - 0.0°C = -14.5°C

2. Calculate m (molality):
m = moles of solute / kilograms of solvent

Since you have the solvent volume in milliliters, convert it to liters:
275 mL = 0.275 L

Now substitute the values into the formula to find the molality:
m = 3.90 mol / 0.275 L = 14.18 mol/kg

3. Calculate the moles of KNO3 using the formula for molality:
m = moles of solute / kg of solvent

Rearrange the formula to solve for moles of solute:
moles of solute = m * kg of solvent

Since the solvent volume is given in liters, you need to convert it to kilograms:
0.275 L * 1 kg/L = 0.275 kg

Now substitute the values into the formula to find the moles of solute:
moles of solute = 14.18 mol/kg * 0.275 kg = 3.90 mol

So, you need 3.90 moles of KNO3 to produce the desired freezing point depression.

To find the mass of KNO3 required, you need to use its molar mass.

4. Find the molar mass of KNO3 (potassium nitrate):
KNO3 = 39.10 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 101.10 g/mol

5. Calculate the mass of KNO3:
mass of KNO3 = moles of KNO3 * molar mass of KNO3

Substitute the values into the formula:
mass of KNO3 = 3.90 mol * 101.10 g/mol = 393.69 g

Therefore, you need to add approximately 393.69 grams of KNO3 to 275 mL of water to produce a solution that freezes at -14.5°C.