If the freezing point of the solution had been incorrectly read 0.3degrees C lower than the true freezing point, would the calculated molar mass of the solute be too high or too low? Explain.

ΔTf = m*Kf*i, solving for molaity,

ΔTf/*Kf*i= m
molality= moles of solute/kg of solution.

Since, 1 mole= molecular weight,
I think decreasing the number of moles decreases the molecular weight.

But I can not be sure about this one.

(1) delta T = i*K*m

(2) m = mols/kg solvent
(3) mols= g/molar mass or
molar mass = g/mols.

T read too low makes delta T too high in 1. That makes m too high.

2. m too high in 2 means mols too high.

3. mols too high in 3 means molar mass too low.
I agree with Devron.

If the freezing point of the solution had been incorrectly read 0.3 degrees Celsius (°C) lower than the true freezing point, the calculated molar mass of the solute would be too low. This is because the observed freezing point depression is directly proportional to the molality of the solute in the solution.

According to Raoult's Law, the freezing point depression (ΔTf) is given by the formula:

ΔTf = Kf * m

Here, Kf is the molal freezing point depression constant of the solvent, and m is the molality of the solute. When the freezing point is measured lower than the true value, the observed ΔTf would be smaller than the actual value.

Since ΔTf is smaller than the actual value, the calculated molality (m) of the solute would also be smaller than the true value. As a result, the calculated molar mass of the solute would appear larger than it actually is.

Therefore, in this scenario, the calculated molar mass of the solute would be too high due to the incorrect reading of the freezing point.

To determine whether the calculated molar mass of the solute would be too high or too low when the freezing point of the solution is incorrectly read lower than the true freezing point, we need to understand the relationship between freezing point depression and molar mass.

Freezing point depression is a colligative property, which means it depends on the number of solute particles rather than their identity. The formula for freezing point depression, ΔTf, is given by:

ΔTf = Kf * m * i

where:
- ΔTf is the freezing point depression
- Kf is the cryoscopic constant, which is characteristic of a particular solvent
- m is the molality of the solute (moles of solute per kilogram of solvent)
- i is the Van't Hoff factor, representing the number of particles into which the solute dissociates (i.e., the number of ions if the solute is ionic or the number of molecules if the solute is molecular).

In this case, if the freezing point of the solution is incorrectly read lower than the true freezing point, it means the measured value of ΔTf would be larger than the actual ΔTf.

Since the freezing point depression is directly proportional to the molality of the solute (m), a higher measured ΔTf would result in a higher calculated molality. The relationship between molality and molar mass (M) can be expressed as:

m = n / (M * solvent mass)

where:
- n is the number of moles of solute
- M is the molar mass of the solute
- solvent mass is the mass of the solvent

If we rearrange the equation, we get:

M = n / (m * solvent mass)

Since the measured molality would be higher due to the incorrect reading, the calculated molar mass (M) would be too low because the numerator (n) remains constant while the denominator (m * solvent mass) becomes larger. Therefore, the calculated molar mass of the solute would be too low.