1.

12 gm of urea is dissolved in 2 liter solution a 300 K temperature. How many gram of NaCl should be dissolved in 10 liter solution so that it becomes iso - osmotic with urea solution ? [At. Wt. of Na = 23, Cl = 35.5 gm/mole]

Answer»

29.25 gm
7.31 gm
5.85 gm
19.5 gm

Solution :`("MOLE of urea")/("volume of urea")=("mole of NACL")/("volume of NaCl")`
`therefore (0.2)/(2)=("mole of NaCl")/(10)`
`therefore` mole of NaCl = 1 mole
But total No. of particles of NaCl = 2
so `("1 mole")/(2)=0.5`mole
WEIGHT of NaCl = mole `xx` weight
`= 0.5xx58.5=29.25` GRAM .


Discussion

No Comment Found

Related InterviewSolutions