AP EAPCET 23 May 2025 Shift 2 Paper

Mole fraction of binary solution,
χ‌glucose ‌+χ‌water ‌=1
χ‌glucose ‌+40×χ‌glucose ‌=1
χ‌glucose ‌=‌

1

41

Assume 1 mole of total solution Moles of glucose
(n‌glucose ‌)=χ‌glucose ‌‌‌×1=‌

1

41

mol
Moles of water (n‌water ‌)=‌

40

41

mol
Mass of glucose ( M‌glucose ‌‌‌ )
‌=n‌glucose ‌‌‌×‌ Molar mass ‌
‌=‌

1

41

×180=‌

180

41

Mass of water (m‌water ‌)=n‌water ‌‌‌× Molar mass
=‌

40

41

×18=‌

720

41

w∕w% of glucose
‌

‌ Mass of glucose ‌

‌ Total mass ‌

×100
=‌ Total mass ‌=‌

180

41

+‌

720

41

=‌

900

41

g
Weight \% of glucose
=‌

‌ 180 41

900 41

×100%=20%