Question

A 7.50g mass of compound A was added to 35.0g of water and it is found that the freezing point has decreased by 3.14°C. What is the molar mass of A?

Answer 1

This problem is providing us with the mass of an unknown compound, the mass of water it was added in and the change (decrease) in the freezing point the solution exhibited. Thus, the problem requires the molar mass of the compound, and at the end, the result turns out to be 126.7 g/mol, according to:

Freezing point depression:

In chemistry, colligative properties are used to understand how the properties of a solvent change as a solute is added, forming a solution. In this case, we can use the following formula, since the solute decreased water's freezing point:

$\Delta T_m=-m*Kf$

Where the molality is the first variable for us to calculate given the other two:

$m=\frac{-3.14\°C}{-1.86\°C/m} =1.69m$

Thus, since this molality is in moles of solute per kilograms of solvent (water), one can find the former by multiplying the 35.0 g of water, but in kilograms:

$n=\frac{1.69mol}{kg} *35.0g*\frac{1kg}{1000g} =0.0591mol$

Finally, we compute the molar mass as required:

$M=\frac{7.50g}{0.0591mol}\\ \\M=126.9g/mol$

As we assumed this solute is nonionizing.

Learn more about colligative properties: brainly.com/question/10323760