Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.
Vapour pressure of water, p1° = 17.535 mm of Hg
Mass of glucose, w2 = 25 g
Mass of water, w1 = 450 g
We know that,
Molar mass of glucose (C6H12O6),
M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
Molar mass of water, M1 = 18 g mol - 1
Then, number of moles of glucose, n1 = 25/180 = 0.139 mol
And, number of moles of water, n2 =450/18 = 25 mol
Now, we know that,
(p1° - p°) / p1° = n1 / n2 + n1
⇒ 17.535 - p° / 17.535 = 0.139 / (0.139+25)
⇒ 17.535 - p1 = 0.097
⇒ p1 = 17.44 mm of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 34: Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhe....
Comments
very good
Did\'nt understand the last second step how does 0.097 comes? Please answer....
Bole to jhakkas
thnx
Irrelevant answer .. Waste of time... Not at all satisfied...