TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

∵ Molar mass of glucose (M)=180g
Molar mass of solute (A), i.e. (MA)=58.5
% of A=117%
% of glucose =7.2%
Also,
∵π (osmotic pressure) =CRT
π=‌

n

V

⋅RT⇒π=‌

w

M

×RT
where, π= osmotic pressure, w=% of sample
M= molar mass, R= gas constant
T= temperature
∵ These are isotonic.
∴‌‌i=‌

MA×%‌ of glucose ‌

%‌ of ‌A×M

i=‌

58.5×7.2

117×180

=‌

421.2

210.6

=2