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3 years ago

WILL GIVE BRAINLIEST!!!

2 answers:

7 0

Hey there!:

ΔTf = Kf * m

Molar mass glucose = 180 g/mol

number of moles glucose:

n = mass of solute / molar mass

n = 21.5 / 180

n = 0.119 moles glucose

Amount of solvent in kg = 255/1000 = 0.255 Kg

Molality = number of moles / solvent

m = 0.119 / 0.255

m = 0.466 moles/kg

Kf for water = - 1.86 ºC/*m

Therefore:

ΔTf = Kf * m

ΔTf = (-1,86) * 0.466

ΔTf = -0.86676 ºC

hope this helps!

max2010maxim [7]3 years ago

7 0

ΔTf = Kf * m Molar mass glucose = 180 g/molthe number of moles glucose:n = mass of solute / molar massn = 21.5 / 180n = 0.119 moles glucoseAmount of solvent in kg = 255/1000 = 0.255 Kg Molality = number of moles / solventm = 0.119 / 0.255 m = 0.466 moles/kgKf for water = - 1.86 ºC/*mTherefore:ΔTf = Kf * m ΔTf = (-1,86) * 0.466ΔTf = -0.86676 ºC

this is to the best of my brain power and i hope this helps with your problem

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$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mol

$\text{Moles of }CO_2=\frac{100.0g}{44g/mol}=2.27mol$

Now we have to calculate the number of oxygen atoms in $H_2O,N_2O,C_3H_6O_2\text{ and }CO_2$

In $H_2O$ :

As, 1 mole of $H_2O$ contains $1\times 6.023\times 10^{23}$ number of oxygen atoms

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As, 1 mole of $N_2O$ contains $1\times 6.023\times 10^{23}$ number of oxygen atoms

So, 2.27 mole of $N_2O$ contains $2.27\times 6.022\times 10^{23}=1.37\times 10^{24}$ number of oxygen atoms

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As, 1 mole of $C_3H_6O_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

So, 1.35 mole of $C_3H_6O_2$ contains $2\times 1.35\times 6.022\times 10^{23}=1.62\times 10^{24}$ number of oxygen atoms

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As, 1 mole of $CO_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

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