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Equal volumes of 30% by mass of H_(2)SO_(4) (density 1.218g mL^(-1)) and of H_(2)SO_(4) (density 1.610 g mL^(-1)) are mixe. If the density of te mixture is 1.452 g mL^(-1) , calculate the molarity oand molality of the solution. |
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Answer» Solution :Calculation of molarity of the solution Let V mL of each sample of `H_(2)SO_(4)` bemixed for 30% `H_(2)SO_(4)` Mass of `H_(2)SO_(4)`=30g ,Mass of solution=100 g `" Volume of solution"=("Mass of soulution")/("Density")= ((100g))/((1.218gmL^(-1)))=100/(1.218)=mL` `100/1.218` mL of sample contains `H_(2)SO_(4)=30 g` V mL of contains `H_(2)SO_(4)=((30g)xx(V mL))/((100//1.18mL))=(0.3654 V)g` For 70% `H_(2)SO_(4)` Mass of `H_(2)SO_(4)`=70g, Mass of solution=100 g `"Volume of solution"=("Mass of solution")/("Density")=((100g))/((1.610gmL^(-1)))=100/(1.610)mL` `100/(1.610)` mL of sample sontains `H_(2)SO_(4)=70g` V mL of sample contains `H_(2)SO_(4)=((70g)(VML))/((100//1.610mL))=(1.127V)g` On mixing the two sample os `H_(2)SO_(4)` Total mass of `H_(2)SO_(4)=0.3654V+1.127V=(1.127V)g` Total volume of solution=2VmL=(0.002V)L `"Molarity of solution(M)"= ("Total mass of" H_(2)SO_(4)//"Molar mass")/("Voluume of solution in litres") ` `=((1.4924 Vg)//(98gmol^(-1)))/((0.002V)L)=7.614 MOL L^(-1)=7.614 M` Calculation of malality of the solution. Total mass of solution= `d xx" volume "=(1.425 gmL^(-1)xx2VmL)=(2.85v)g` Total mass of ACID =(1.4924 V)g Mass of WATER = `(2.85 V-1.4924 V) g=(1.3576xx10^(-3)V) kg` `"Molality of solutin (m)"=("Mass of" H_(2)SO_(4)//"Molar mass")/("Mass of SOLVENT in kg")=((1.4924V)g//(98gmol^(-1)))/((1.3576xx10^(-3)V)kg)` =11.217 mol `kg ^(-1)`=11.217 m |
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