An aqueous solution was made by dissolving 72.9 grams of glucose, C6H12O6, into 115 grams of water. The vapor pressure of the pure water is 26.4 Torr. The vapor pressure of water over this solution is: (a) 27.9 Torr (b) 24.1 Torr (c) 26.8 Torr (d) 24.8 Torr PLease answer this as quick as possible

The correct answer is (d) 24.8 Torr

Explanation:

When a solute is added to a solvent, the water pressure of the solution is lower than the vapor pressure of the pure solvent. This is called vapor pressure lowering and it is given by the following expression:

Psolution= Xsolvent x Pºsolvent

We have to calculate Xsolvent (mole fraction of solvent) which is given by the number of moles of solute divided into the total number of moles.

First, we calculate the number of moles of solute and solvent. The solute is glucose (C₆H₁₂O₆), and its number of moles is calculated from the mass and the molecular weight (MM):

MM (C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

moles of glucose= mass/MM= (72.9 g)/)(180 g/mol)= 0.405 moles

The solvent is water (H₂O) and again we calculate the number of moles as follows:

MM(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol

moles of water= mass/MM= (115 g)/(18 g/mol)= 6.389 moles

Now, we calculate the total number of moles (nt):

nt= moles of glucose + moles of water= 0.405 moles + 6.389 moles= 6.794 moles

The mole fraction of water (Xsolvent) is given by:

Xsolvent= moles of water/nt= 6.389 moles/6.794 moles= 0.940

Finally, the vapor pressure of water over the solution will be the following:

Psolvent= Xsolvent x Pºsolvent= 0.940 x 26.4 Torr= 24.8 Torr

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

the dependant variable is spilled water

the independent is the heat