Question

A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . 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

Answer 1

Answer : The correct answer for molarity = 0.19 $\frac{mol}{L}$ and Molality = 0.18 $\frac{mol}{Kg}$.

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

$Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}$\

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

$Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}$

Molar mass of aniline (C₆H₅NH₂) = 93.13 $\frac{g}{1 mol}$

Mass of aniline = 6.2 g

Plugging values in mole formula :

$Mole = \frac{6.2 g}{93.13 \frac{g}{1 mol}}$

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

$Volume of solution = \frac{350 mL}{1000mL} * 1 L$

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

$Molarity = \frac{0.0665 mol}{0.350 L }$

Molarity = 0.19 M or 0.19 $\frac{mol}{L}$

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2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

$Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}$

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

$Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}$

Plugging value in density formula :

$1.04 \frac{g}{mL} = \frac{ mass }{350 mL}$

Multiplying both side by 350 mL

$1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL$

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

$Mass of solvent = \frac{364 g}{1000g} * 1 Kg$

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

$Molality = \frac{0.0665 mol}{0.364 Kg}$

Molality = 0.18 m or 0.18 $\frac{mol}{Kg}$