18 g of glucose (C₆H₁₂O₆, M = 180 g/mol) is dissolved in 1 kg of water — JEE Chemistry
18 g of glucose ($C_6H_{12}O_6$, $M = 180$ g/mol) is dissolved in 1 kg of water. Calculate: (a) molality, (b) boiling point elevation, (c) osmotic pressure at 300 K. ($K_b = 0.52$ K·kg/mol, $R = 0.0821$ L·atm/mol·K)
1 Answer
AAayushaRai25
✓ Accepted
· 1mo ago
▲ 4
$$m = \frac{18/180}{1} = 0.1 mol/kg$$
(a) $m = 0.1 molal$
(b) $\Delta T_b = K_b \times m = 0.52 \times 0.1 = 0.052 K$
(c) Molarity $\approx 0.1$ mol/L (dilute solution):
$$\pi = MRT = 0.1 \times 0.0821 \times 300 = 2.463 atm$$
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Discussion (2)
S
This finally made it click for me — thank you!
S
How do we know the approximation is valid here?