Question

an aqueous solution is prepared by diluting 3.30mL acetone(d=0.789g/mL) with water to final volume of 75.0mL.The density of tHe solution is 0.993g/mL. what is the molality ,molality of acetone in the solution?

Answer 1

$\bf\huge\underline{\underline{ Given :- }}$

• an aqueous solution is prepared by diluting 3.30mL acetone(d=0.789g/mL) with water

• final volume of 75.0mL.The density of tHe solution is 0.993g/mL.

$\bf\huge\underline{\underline{ To \: Find:- }}$

• molality ,
• molality of acetone in the solution

$\bf\huge\underline{\underline{ Answer :- }}$

$\bf\red{ A \implies H_{2}O }$

$\bf\red{ B \implies CH_{3} CO CH_{3} }$

• density = $\sf \frac{ Mass }{ Volume }$

• Mass = density × Volume

$\rule{10cm}{0.0005cm}$

• $\sf \rho_{B}$ = 0.789 g/mL

• $\sf V_{B}$ = 3.30 mL

• $\sf W_{B}$ = $\sf \rho_{B} × V_{B}$

• $\red{\implies}$ $\sf W_{B}$ = 0.789 × 3.30

• $\red{\implies}$ $\sf W_{B}$ = 2.60 g

$\rule{10cm}{0.0005cm}$

• mass of solution = 0.993 × 75

• $\red{\implies}$ = 74.475 g

• $\sf W_{A} = 74.475 - 2.60$

• $\sf W_{A} = 71.87 g$

$\rule{10cm}{0.0005cm}$

• Molecular Mass of B = 58 g/mol

• no. of moles of B = $\sf \frac{W_{B}}{{M_{B}}}$

• $\red{\implies}$ no. of moles = $\sf \frac{ 2.60 }{ 58 }$

• $\red{\implies}$ no. of moles = 0.0448 moles

$\rule{10cm}{0.0005cm}$

Molarity

• $\boxed{\bf\red{ Molarity = \frac{ no.\: of \: moles }{ L of solvent }}}$

• $\red{\implies}$ Molarity = $\frac{0.0448}{0.0750}$

• $\red{\implies}$ Molarity = 0.598 M

$\rule{10cm}{0.005cm}$

Molality

• $\boxed{ \bf\red{ Molality = \frac{no.\: of \: moles }{ Kg \: of \: solvent }}}$

• $\red{\implies} \sf Molality = \frac{ 0.0448}{0.0718}$

• $\red{\implies}$ Molality = 0.624 m

$\rule{10cm}{0.05cm}$