How many grams of ammonium nitrate nedd to be added to 500g of water so that the freezing point becomes -8.3 °C?

-8.3=-.52*molesammonuimNO3/.5

gramsNH4NO3=molmass*8.3/.52*.5
look up the formula mass of ammonium nitrate, and solve.

Can you please give me the formula

To determine the grams of ammonium nitrate needed to lower the freezing point of water to -8.3 °C, we need to use the formula for calculating the freezing point depression:

ΔT = K_f × m

Where:
ΔT is the change in freezing point
K_f is the cryoscopic constant for water (-1.86 °C/m)
m is the molality of the solute (moles of solute per kg of solvent)

First, let's calculate the change in freezing point (ΔT):

ΔT = (desired freezing point) - (freezing point of pure water)
ΔT = (-8.3 °C) - (0 °C)
ΔT = -8.3 °C

Next, we need to calculate the molality (m) of the ammonium nitrate. Molality is defined as the moles of solute divided by the mass of the solvent, in this case, water.

molality (m) = moles of solute / mass of solvent

Given that we have 500 grams of water, we need to convert it to kilograms:

mass of solvent = 500 g = 0.5 kg

Now, we need to convert ΔT from °C to Kelvin (K). The two units are the same, so:

ΔT = -8.3 °C = -8.3 K

Now, we can rearrange the formula to solve for m:

m = ΔT / K_f

m = -8.3 K / (-1.86 °C/m)

m ≈ 4.4628 mol/kg (rounded)

Finally, to find the number of moles of solute required, we need the molecular weight of ammonium nitrate. The molecular weight of ammonium nitrate is approximately 80.04 g/mol.

m = moles of solute / mass of solvent

4.4628 mol/kg = moles of solute / 0.5 kg

moles of solute ≈ 4.4628 mol/kg × 0.5 kg

moles of solute ≈ 2.2314 mol

Finally, multiply the moles of solute by the molecular weight to obtain the mass of ammonium nitrate required:

mass of solute = moles of solute × molecular weight

mass of solute ≈ 2.2314 mol × 80.04 g/mol

mass of solute ≈ 178.16 g

Therefore, you would need approximately 178.16 grams of ammonium nitrate to be added to 500 grams of water for the freezing point to become -8.3 °C.