6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to
A. 0.0196
B. 0.5
C. 0.1
D. 0.0202

8 views

1 Answers

Correct Answer - A
`(P_(1)^(o)-P)/P_(1)^(o)=("Mass of solute/molar mass")/("Mass of soute"/"Molar mass"+"Mass of solvent"/"Molar Mass")`
`(6//60)/(6/60+90/18)=(1//10)/(1/10+5/1)=1/51=0.0196`

8 views

Related Questions