Calculate the amount of CaCl_(2) (van't Hoff factor I =2*47) dissolved in 2*5 L solution so that its osmotic pressure at 300 K is 0*75 atmosphere. Given : Molar mass of CaCl_(2) is 111 g."mol"^(-1), R=0*082L." atm "K^(-1)"mol^(-1).

Calculate the amount of CaCl_(2) (van't Hoff factor I =2*47) dissolved

1.	Calculate the amount of CaCl_(2) (van't Hoff factor I =247) dissolved in 25 L solution so that its osmotic pressure at 300 K is 075 atmosphere. Given : Molar mass of CaCl_(2) is 111 g."mol"^(-1), R=0082L." atm "K^(-1)"mol^(-1).
Answer» Solution :The osmostic pressue (calculated), `pi=(n)/(v)RT` . . .(i) where, V=volume `=25L,` T=Temperature = 300 K, n= dissolving moles of `CaCl_(2)` `=("Wt. of solute(g)")/("Molecular mass of solute (M)")` Thus equation (i) becomes `pi_("Cal")=(g)/(MV)RT` `:,""pi_("Cal")=(g)/(111"g mol"^(-1)xx25l)xx0082l" atm"` `K^(-1)"mol"^(-1)xx300K` `=(246)/(2775)atm=08=886xxg" atm. gm"` Van's HOFF factor (i) `=("observed osmotic pressure "(pi_(ob)))/("calculated osmotic pressure"(pi_(Cal)))` `:.""247=(075" atm")/("calculated osmotic pressure")` `:.""247=(075" atm")/(pi_"Cal")or,pi_("Cal")=03036` atm Thus, `03036=00886xx"g gm or",g=34271" gm"` Hence, the amount of `CaCl_(2)` dissolve in `25" l"` solution is `3*4271` gm.