SELECT * FROM question_mgmt as q WHERE id=1777 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Solutions boiling point of water at 750 mm hg is 99 63 deg c

Question:

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.

Answer:

Here, elevation of boiling point ΔTb= (100 + 273) - (99.63 + 273)

= 0.37 K

Mass of water, wl = 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:

= (0.37 x 342 x 500) /  (0.52 x 1000)

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.


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SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='9' AND ex_no!=0 AND status=1 and id!=1777 ORDER BY last_viewed_on desc limit 0,10

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