Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given. {:("Column I","Column II"),("For a 5% solution of "H_(2)SO_(4)(d ="1.01 g mL"^(-1)),),("(A) Molarity of the solution","(p) 0.537"),("(B) Molality of the solution","(q) 0.0096"),("(C) Mole fraction of "H_(2)SO_(4),"(r) 0.05"),("(D) Mass fraction of "H_(2)SO_(4),"(s) 0.515"):}

Match the entries of column I with appropriate entries of column II an

1.	Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given. {:("Column I","Column II"),("For a 5% solution of "H_(2)SO_(4)(d ="1.01 g mL"^(-1)),),("(A) Molarity of the solution","(p) 0.537"),("(B) Molality of the solution","(q) 0.0096"),("(C) Mole fraction of "H_(2)SO_(4),"(r) 0.05"),("(D) Mass fraction of "H_(2)SO_(4),"(s) 0.515"):}
Answer» A-r, B-s, C-p, D-q A-q, B-p, C-s, D-r A-s, B-p, C-q, D-r A-s, B-r, C-p, D-q Solution :`%% H_(2)SO_(4)` solution MEANS 5 g `H_(2)SO_(4)` in 100 g solution, i.e., SOLVENT `(H_(2)O)=95g=0.095kg` As `d=1.01" g mL"^(-1)`, volume of 100 g solution `=(100)/(1.01)mL=99mL=0.099L` `"Molarity "=("5/98 MOL")/("0.099 L")="0.515 M"` Mole fraction `=(5//98)/(5//98+95//18)=(0.051)/(0.051+5.278)=0.0096` Mass fraction `=(5)/(100)=0.05`.