Question

A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the volume of the solvent does not change when the aniline dissolves in it.
Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

How do I enter a number in scientific notation?

a. molarity =
b. molality =

Answer 1

Answer:

a. Molarity= $M =2.1x10^{-1}M$

b. Molality= $m=2.0x10^{-1}m$

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

$M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M$

Moreover, the molality:

$m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m$

Best regards.