Match the following :
`{:("Column-I","Column-II"),((A)"4.5 m solution of"CaCO_3"density 1.45 gm/ml",(p)"Mole fraction of solute is 0.2"),((B)"3 M 100 ml" H_2SO_4 "mixed with 1 M 300 ml"H_2SO_4"solution",(q)"Mass of the solute is 360 gm"),(( C)"14.5 m solution of Ca",(r)"Molarity=4.5"),((D)"In 2 litre solution of 4 M NaOH, 40 gm NaOH is added",(s)"Molarity 1.5"),((E)"5m (molal)NaOH solution",(t)"16.66%(w/w) of NaOH solution"):}`

Answer

A-r , B-r , C-p D-q,r E-t

Explanation

(A)4.5 m,`CaCO_3` means 4.5 moles of `CaCO_3` is present in 1000 g solvent.
mass of solute =4.5x100=450 g
mass of solution =1450 g. So, volume of solution =`1450/1.45=1000 ml`
Hence molarity=4.5 M
(B)Resultant molarity =`(3xx100+1xx300)/400=3/2=1.5 M`
( C)mole fraction `=14.5/(14.5+55.5)=0.2`
(D)moles of NaOH in 2 ltr =4x2=8 mole
moles of NaOH added `=40/40=1` mole
`:.` Molarity =`9/2=4.5` M
mass of NaOH=9x40=360 g
(E) % (w/w) NaOH `=(5xx40)/(5xx40+1000)xx100=16.66%`