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ololo11 [35]
3 years ago
8

A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0 g of water. Calculate the boiling point of the solution

. Urea is a nonelectrolyte.
Chemistry
1 answer:
andrew11 [14]3 years ago
5 0

<u>Answer:</u> The boiling point of solution is 101.56°C

<u>Explanation:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure water = 100°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point elevation constant = 0.52°C/m.g

m_{solute} = Given mass of solute (urea) = 27.0 g

M_{solute} = Molar mass of solute (urea) = 60 g/mol

W_{solvent} = Mass of solvent (water) = 150.0 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=1\times 0.52^oC/m\times \frac{27\times 1000}{60\times 150}\\\\\text{Boiling point of solution}=101.56^oC

Hence, the boiling point of solution is 101.56°C

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3 years ago
Express the following in liters at STP:<br><br><br><br> 6.62 x 10-3 moles HF
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Answer:

The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L

Explanation:

Given data:

Number of moles of HF =  6.62×10⁻³ mol

Volume of HF in litter at STP = ?

Solution:

The given problem will be solve by using general gas equation,

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Standard pressure = 1 atm

Now we will put the values in formula.

1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K   × 273 K

V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K   × 273 K  / 1 atm

V = 148.38×10⁻³ L

Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.

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