an aqueous solution containing 17.5g of an unknown molecular(nonelectrolyte) compound in 100.0g of water has a freezing point of 1.8 degress C. calculate the molar mass of the unknown compound

im lost where did you get the 52 from

I think Mr. Pursley typed in the boiling point elevation constant of 0.52 C when he intended to type in 1.86 for the freezing point constant.

not sure where to start. do i use the freezing point depression equation

To calculate the molar mass of the unknown compound, we can use the concept of freezing point depression, which is the difference between the freezing point of a pure solvent (in this case, water) and the freezing point of the solution.

The freezing point depression (∆Tf) can be calculated using the formula:

∆Tf = Kf * molality

where Kf is the cryoscopic constant for the solvent (water), and molality is the concentration of the solute in moles per kilogram of solvent.

First, we need to determine the molality of the solution:

Molality (m) = moles of solute / mass of solvent (in kg)

The mass of the solvent (water) is given as 100.0g, but we need to convert it to kilograms:

Mass of solvent = 100.0g = 0.100 kg

To calculate the moles of solute, we will use the formula:

moles of solute = mass of solute / molar mass

The mass of the solute is given as 17.5g. Plugging these values into the equation, we have:

moles of solute = 17.5g / molar mass

Now, we substitute these values into the formula for molality:

m = moles of solute / mass of solvent = (17.5g / molar mass) / 0.100 kg

Next, we substitute the value of molality into the equation for freezing point depression:

∆Tf = Kf * molality

The cryoscopic constant for water (Kf) is 1.86°C/m. The freezing point depression (∆Tf) is given as 1.8 degrees Celsius.

1.8 = (1.86°C/m) * [(17.5g / molar mass) / 0.100 kg]

Simplifying the equation, we can solve for the molar mass:

molar mass = (1.86°C/m) * [(17.5g / 0.100 kg) / 1.8]

Calculating this expression will give you the molar mass of the unknown compound.

Really? 1.8C, above the freezing point of water? I bet you mean -1.8C.

-1.8C=-.52*17.5/(molmass*.100)

solve for molmass