4 g of na2co3.xh2o is dissolved in 200 ml of solution. 20 ml of this solution requires 19.8 ml of 0.1 n hcl for neutralisation. the value of x is
Answers
Explanation:
Na2CO3.XH2O+2HCl→2NaCl+(X+1)H2O+CO2
Na2CO3.XH2O+2HCl→2NaCl+(X+1)H2O+CO2The strength of a solution is defined as the amount of solute in grams, present in one litre of solution. It is expressed in gL−1. We are given that 0.7gofNa2CO3.XH2O is dissolved 100 ml of solution.
Na2CO3.XH2O+2HCl→2NaCl+(X+1)H2O+CO2The strength of a solution is defined as the amount of solute in grams, present in one litre of solution. It is expressed in gL−1. We are given that 0.7gofNa2CO3.XH2O is dissolved 100 ml of solution. The strength of the solution is therefore =0.7/[100/1000]=7gL−1
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl M2-> molarity of HCl
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl M2-> molarity of HCl V2-> volume of HCl used
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl M2-> molarity of HCl V2-> volume of HCl used Since HCl is monobasic, therefore its molarity is the same as its normality. Substituting the given values in equation (1), we get
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl M2-> molarity of HCl V2-> volume of HCl used Since HCl is monobasic, therefore its molarity is the same as its normality. Substituting the given values in equation (1), we get2×M1×20=1×0.1×19.8
=7gL−1The molarity equation is written as n1×M1×V1=n2×M2×V2....(1)where n1-> acidity of Na2CO3 M1-> molarity of Na2CO3 V1-> volume of sodium carbonate solution used n2-> basicity of HCl M2-> molarity of HCl V2-> volume of HCl used Since HCl is monobasic, therefore its molarity is the same as its normality. Substituting the given values in equation (1), we get2×M1×20=1×0.1×19.8Therefore M1=19.8/400=0.0495M
=0.0495MNow Molarity= StrengthingL−1/molar mass of solute
=0.0495MNow Molarity= StrengthingL−1/molar mass of soluteTherefore, molar mass of Na2CO3.xH2O =70.0495=141.414 g/mol
=141.414 g/mol But molar mass of Na2CO3.xH2O= Mass of anhydrous Na2CO3+ mass of x molecules of water =106+18x
=141.414 g/mol But molar mass of Na2CO3.xH2O= Mass of anhydrous Na2CO3+ mass of x molecules of water =106+18xTherefore 106 + 18x = 141.414 , which gives x=35.414/18=1.976
=1.976Hence the value of x here can be rounded off to 2.
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