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nikklg [1K]
3 years ago
6

A 3.50 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 mL teapot containing 80∘ C water (density of water at 80∘ C= 0.97

5 g/mL). What is the molality of the sugar solution?
Chemistry
2 answers:
KIM [24]3 years ago
6 0

Answer: 0.04893mole/kg

Explanation:

Molality is defined as the mole of solute per kilogram of the solvent.

Sugar is the solute while water is the solvent.

Molar mass of sugar =

1*12+1*22+11*16

12+22+176=210g/mol.

Mole of solute= reacting mass /molar mass

Reacting mass 3.50g and molar mass 210g/mol

Mole= 3.50/210=0.0167moles.

For solvent

Density=mass/volume

Mass= density *volume

0.975*350

341.25g

Converting gram back to kilogram

341.25/1000= 0.34125kg

Molality =0.0167/0.34125

=0.04893mole/kg

mestny [16]3 years ago
6 0

Answer:

Molality is 0.03 m

Explanation:

Molality indicates the moles of solute that are contained in 1000g of solvent.

We have solvent volume and density, so let's find out the mass:

Solvent density = Solvent mass / Solvent volume

Solvent density . Solvent volume = Solvent mass → 0.975 g/mL . 350 mL

Solvent mass = 341.25g

Now we need the moles of solute → mass of solute / molar mass

3.50 g / 342 g/mol = 0.0102 moles

So finally we can make a rule of three:

341.25 g of solute contain 0.0102 moles of solute

1000 g of solute, may contain (1000 . 0.0102) / 341.25 = 0.03 m

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                =   374.90 kPa 

Calculation:
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So,
                                    3.70 atm  =   X
Solving for X,
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How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +
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Explanation:

To calculate the moles :

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\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles

The balanced chemical reaction is:

4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)

According to stoichiometry :

7 moles of O_2 produce =  6 moles of H_2O

Thus 0.900 moles of O_2 will produce =\frac{6}{7}\times 0.900=0.771moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g

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<span>Multiply both sides by 1.50 L to isolate moles of solute on the right. </span>

<span>8.1 mol = moles of solute </span>


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