Calculate the boiling point of solution when 4g of Mg `SO_(4) (M=120g "mol"^(-1))` was dissolved in 100g of water, assuming `MgSO_(4)` undergoes complete ionization
`(K_(b) " for water " = 0.52 K " kg mol"^(-1))`
`DeltaT_(b) = iK_(b) xx m`
`T_(S)- Ti = (i xx 0.52 xx WB)/(M_(g) xx W_(A)(Kg))`
`i=2, M_(B) = 120 g//"mol" W_(B) = 4g, W_(A) = 100`
`T_(S) - 100= (2 xx 0.52 xx 4 xx 1000)/(120 xx 100)`
`T_(S)-100 = 0.346`
`T_(S) = 100+0.346`
`T_(S) = 100.346 ""^(@)C`
Total mass of the water `=M=100g`
Latent heat of vaporization of water at `0^(@)C`
`=L_(1)=21.0xx10^(5)J//Kg`
& Latent heat of fusion of ice `=L_(2)=3.36xx10^(5)J//kg`
Suppose, the mass of the ice formed...
The desired equation is
`BaCl_(2)+underset(=174g)underset(2xx39+32+64)(K_(2)SO_(4)) to underset(=223g)underset(137+32+64)(BaSO_(4))+2KCl`
`233g of BaSO_(4)` obtained from 174g of `K_(2)SO_(4)`
1.2 of `BaSO_(4)` will be obtained from `(174)/(233)xx1.2`
`=0.8961"g of "K_(2)SO_(4)`
So, 0.891g of `K_(2)SO_(4)`...
`40ppm=("Mass of" SO_(2))/("Mass of air")xx10^(6)rArr"Mass of" SO_(2)=(40xx"Mass of air")/(10^(6))`
`d_("air")=("Mass of air")/(V_("air")(lit.))=(2gm)/(lit.)rArr"Mass of air"=2xx100=200gm`
Mass of `SO_(2)=(400xx200)/(10^(6))=8xx10^(-3)gmrArr 8mg`
Assume that mole of product mixture =x mole
`n_(a)=50=xxx0.6+x xx0.2-0.8x`
`x=(50)/(0.8)=(500)/(8)=62.5`
`n_(o_(2))` remain=62.5xx0.2-12.5
`n_(o)=2xxn_(SO_(2))+3xxn_(SO_(3))=2xx x xx0.6+3xx x xx0.2`
`n_(o)=1.2x+0.6x=1.8x=1.8xx62.5`
`n_(o_(2))=(1.8xx65)/(2)=0.9xx65=56.25`
`n_(o_(2))` total require `=56.25+12.5=68.75`