<span>B) phosphodiester </span> is the correct answer
The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
We have get the mass of gaseous water after evaporation in a closed container.
The mass of water vapor after evaporation is 5 grams.
In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.
Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.
Randomness of molecules increases in gaseous state than liquid state.
The freezing point depression of the solution or pure substance that is added with the solvent is calculated through the equation,
ΔTf = Kfm
where ΔT is the freezing point depression, Kf is the constant for water given to be -1.86°C/m and m is the molality of the solution.
Molality is calculated through the equation,
m = number of moles solute/ kg of solvent
Calculation of molality is shown below.
m = (21.5 g C6H12O6)(1 mol/180 g) / (0.255 kg)
m = 0.468 molal
The freezing point depression is then,
ΔTf = (-1.86°C/m)(0.468 m) = -0.87°C
<em>Answer: -0.87°C</em>
I’m pretty sure the answer is true
Have a nice day! :D
