JEE Advanced Model Paper 18 with solutions for online practice

π = mRT (m = molarity, R = 0.082 L atm/mol. K, T = 25°C = 298K )
M = m =

π

RT

=

5×10−3

0.082×298

= 2 × 10−4
M =

w

m×V

m =

w

M×V

m =

10

2×10−4

= 5 × 104
Each glucose unit (mw = 180)
Bonded to the starch chain with bss of 1 H2O (mm = 18) giving a effective molecular mass = 180 - 18 = 162
So approximate average number of glucose =

5×104

162

= 309 ≅ 300