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प्रश्न
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? Molal elevation constant for water is 0.52 K kg mol−1.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C?
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उत्तर
Here, elevation of boiling point (ΔTb) = (100 + 273) − (99.63 + 273)
= 0.37 K
Mass of water (w1) = 500 g
Molar mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16
= 342 g mol−1
Molal elevation constant (Kb) = 0.52 K kg mol−1
We know that,
ΔTb = `(K_b xx 1000 xx w_2)/(M_2 xx w_1)`
⇒ w2 = `(Delta T_b xx M_2 xx w_1)/(K_b xx 1000)`
= `(0.37 xx 342 xx 500)/(0.52 xx 1000)`
= 121.67 g (approximately)
= 122 g
Hence, 122 g of sucrose is to be added.
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