Calculate the mass in grams for 0.251 moles of Na2CO3

The boiling point of water is 100.0°C at 1 atmosphere. How many grams of sodium acetate (82.04 g/mol), must be dissolved in 283.

Alecsey [184]

Answer:

I have to weight 10,04 g of acetate sodium

Explanation:

This is the colligative propertie about elevation of boiling point

ΔT = Kb . m . i

ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure - we have this data 0,450°C

Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)

m means molality (moles of solute in 1kg of solvent)

i means theVan 't Hoff factor (degree of dissociation for a compound)

For the sodium acetate is 2

NaCH3COO ---> Na+ + CH3COO-

0,450°C = 0,52°C.kg/m . m . 2

0,450°C / (0,52°C.kg/m . 2) = m

0,432 m/kg = m

This number means I have 0,432 moles of acetate sodium, my solute in 1kg of water, my solvent. But I don't have 1000 g (1kg) I only have 283 g so let's make the rule of three:

1000 g _____ 0,432 moles

283 g ______ (283g .0,432m) / 1000g = 0,122 moles

Now that I have the moles of acetate sodium, I have to find the mass.

Moles . molar mass = mass

0,122 moles . 82.04 g/mol = 10,04 g

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3 years ago

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How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules