A student dissolves of aniline in of a solvent with a density of . 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.

Molarity: 0.21M

Molality: 0.20m

Explanation:

...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...

To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):

Moles aniline:

Molar mass:

6C: 6* 12.01g/mol = 72.06g/mol

7H: 7*1.008g/mol = 7.056g/mol

N: 1*14.007g/mol = 14.007g/mol

72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol

Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles

Liters solution:

200mL * (1L / 1000mL) = 0.200L

kg solvent:

200mL * (1.05g/mL) * (1kg/1000g) = 0.210L

Molarity:

0.04188mol / 0.200L = 0.21M

Molality:

0.04188mol / 0.210L =0.20m