How many moles are in 1.5x1023 molecules of Na2SO4?
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Answer : The correct answer for molarity = 0.19 and Molality = 0.18
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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