Question

A student dissolves of 15 g 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.

Answer 1

The given question is incomplete. The complete question is ;

A student dissolves of 15 g aniline 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.

Answer: The molarity is 0.81 M and molality is 0.82 m

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 200 ml

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.16\times 1000}{200}=0.81M$

Thus molarity is 0.81 M

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{15g}{93g/mol}=0.16mol$

Mass of solution = $Density\times Volume=1.05g/ml\times 200ml=210g$

mass of solvent = mass of solution - mass of solute = (210 - 15) g = 195 g

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.16\times 1000}{195g}=0.82mole/kg$

Therefore, the molality of solution is 0.82m