Astudent 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 : The correct answer for molarity = 0.19 and Molality = 0.18 .

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 :

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Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

Molar mass of aniline (C₆H₅NH₂) = 93.13

Mass of aniline = 6.2 g

Plugging values in mole formula :

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 = 0.350 L

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

Molarity = 0.19 M or 0.19

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

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 :

Plugging value in density formula :

Multiplying both side by 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 = 0.364 Kg

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

Molality = 0.18 m or 0.18