Question

Mole fraction of urea in 900 gram water is 0.05 density of solution is `1.2g//cm^(3)`. Find the molarity of solution.

Answer 1

Correct Answer - 2.98 or 2.99
`wt_(H_(2)O)=900gimpliesn_(H_(2)O)=(900)/(18)=50`
Let moles of urea =y
`impliesx_(urea)=(y)/(y+50)=0.05`
`impliesy=(50)/(19)`
`thereforeWt_(urea)=((50)/(19))xx60=(3000)/(19)`
`thereforeWt_(solution)=(3000)/(19)+900=((201))/(19)xx100`
`rho_("solution")=1.2g//cm^(3)`
`impliesV_("solution")=(wt)/(rho)=((201)/(19))xx(100)/(1.2)`
`impliesM_(urea)=(n_(urea))/(V_(ml))xx1000=(((50)/(19))xx1000)/(((201)/(19))xx(100)/(1.2))=((500)(1.2))/(201)=(600)/(201)cong2.985`
Answer after rounding off =2.98