125 g of the non-volatile solute glucose, GH O,is dissolved in 125 g of water at

Question

125 g of the non-volatile solute glucose, GH O,is dissolved in 125 g of water at 25.0°C. IF che vapor pte water at 25.0'C is 23.7 Tor, what is the vapor pressure o caloaare the Osmobdc presure at S0 degrees Cekius of a placose solution G,H,O, that has 60 grams of glucose dissolved in enough water to make 1500 ml answer: 5.89 atm What is the expected freezing point of a 0.50 m solution of NazCOs in water? Ke for water is 1,86 C/m. -2.8°C A concentration-time study of the gas phase reaction 2 A, 3 A, produced the data in the table below. A concentration-time study of the gas phase reaction 2 A,- 3 A produced the data in the table below Time ()[AJ M) 10 20 30 4.00 × 104 2.00× 10" 1.00× 10" 5.00 ×105 [AJ (M) 0 3.00 x 10 4.50 × 10" What is the average rate of formation of A, in the time interval 20-30 seconds? Answer: 7.50 × 10-6 M/s The following set of data was obtained by the method of initial rates for the reaction: (H,C),CBr + OH- (H,C),COH + Br. What is the order of reaction with respect to ion, OH-? [OHI M Initial Rate (M/s) [(H,C)CBr) M 0.25 0.50 0.50 0.25 0.25 0.50 1.1 × 10" 2.2× 10" 22 x 10 Answer: zero Dang4

Explanation / Answer

125 g glucose in 125 g water

Given,

Moles of glucose = 125 g/180.156 g/mol = 0.694 moles

moles of water = 125 g/18.015 g/mol = 6.94 moles

Total moles = 7.634 moles

moles fraction of water = 6.94/7.634 = 0.91

vapor pressure of solution = 0.91 x 23.7 torr = 21.57 torr

===

Osmotic pressure = iMRT

Van't hoff factor (i) = 1

molarity of solution (M) = 60 g/180.156 x 1.5 L = 0.222 M

R = gas constant

T = 50 oC + 273 = 323 K

So,

osmotic pressure = 1 x 0.222 x 0.0821 x 323

                            = 5.89 atm

===

Expected freezing point = -iKfm

Van't hoff fator (i) = 3 [2Na+ + 1CO3^2- ions]

m = 0.50 m

Kf = 1.86 oC/m

So,

Expected freezing point = -3 x 1.86 x 0.50

                                       = -2.8 oC

===

From the given reaction data,

average rate of formation between 20 to 30 s

= [(1 x 10^-4 - 5 x 10^-5)/10](3/2)

= 7.5 x 10^-6 M/s

===

From the tert-butyl bromide data.

from 2nd and 3rd data points

when concentration of [(CH3)3CBr is constant, [OH-] is doubled,

the rate remained the same

So,

order with respect to [OH-] = zero.

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