Question

What mass of non volatile solute urea needs to be dissolved in 100g of water?

Answer 1

Hey mate here's your answer

Hello Student,

Po– P/Po = w/3 /w/m+W/M

Here, w and m are wt. and molecular wt. of solute, W and M are wt. and molecular weight of solvent

p = pressure of solution;

po = Normal vapour pressure

Let the initial (normal) pressure (po) = p

∴ Pressure of solution = 75/100 * p = 3/4 p

M = 60, M = 118, W = 100 gm

∴ p-3/4 p/p = w/60 /w/60 + 100/18

1/4 = w/60 /(w/60) + 5.55

Or 4w/60 = w/60 + 5.55

3w/60 = w/20 = 5.55

Or w = 111 g

Molality = No. of moles of solute/Wt. of solvent * 1000

= 111 *1000/60 *100 = 18.52 m

Thanks

Answer 2

p°_p/p°= w/3 /w/m +2/M

ĤĔŔĔ ,Ŵ ÁŃĎ m ÁŔĔ ŴĔĨĞĤŤ ÁŃĎ MŐĹĔČÚĹÁŔ ŴĔĨght ŐŦ ŚŐĹÚŤe ,Ŵ ÁŃĎ M ÁŔĔ ŴĔĨĞĤŤ ÁŃĎ MŐĹĔČÚĹÁŔ ŴĔĨĞĤŤ ŐŦ ŚŐĹVĔŃŤ.

P=PŔĔŚŚÚŔĔ ŐŦ ŚŐĹÚŤĨŐŃ

P°= ŃŐŔMÁĹ VÁPŐÚŔ PŔĔŚŚÚŔĔ .

ĹĔŤ ŤĤĔ ĨŃĨŤĨÁĹ ( ŃŐŔMÁĹ ) PŔĔŚŚÚŔĔ (P°)=P .

ŤĤĔŔĔŦŐŔĔ PŔĔŚŚÚŔĔ ŐŦ ŚŐĹÚŤĨŐŃ= 75 /1000× P ×p =3/4p .

M=60, M=80,Ŵ=100gm

P-3/4 P/p = w/60/W/60+100/18

1/4=w/60/(w/60)+5.55

or 4w/60=w/60+5.55

3w/60=w/20 = 5.55 or w=111g

molarity =no.of moles of solute /weight of solvent ×1000

=111×1000/60=18.52m.

ĤŐPĔ ŤĤĨŚ ÁŃŚŴĔŔ ĤĔĹPŚ ŶŐÚ .ĨŦ ŚŐ PĹĔÁŚĔ MÁŔĶ MĔ ÁŚ ßŔÁĨŃĹĨĔŚŤ