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# A solution containg 1 gm per dm3 of urea is isotonic with 5% solution of a nonvolatile solute. What is the molecular mass of this non volatile solute

Vikas TU
14149 Points
4 years ago
Given information :
Urea arrangement is 1gm for every dm3, so m=1g and V = 1000 ml
Non unstable arrangement is 5%, so m=5g and V = 100ml
General Gas condition
PV = nRT      PV =m/M RT
PM = m/V RT
For urea P1 = (m_1 RT)/(M_1 V_1 )
For Non unstable Solute P2 = (m_2 RT)/(M_2 V_2 )
Since both arrangements are isotonic, so P1 = P2
Subsequently we can compose
(m_1 RT)/(M_1 V_1 )=(m_2 RT)/(M_2 V_2 )
m_1/(M_1 V_1 )=m_2/(M_2 V_2 )
1g/(60 g/mol X 1000ml)=5g/(M_2 X100ml)
M_2=3000 g/mol
In this way, atomic weight of non unstable solute is 3000 g/mol.