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Marizza181 [45]
1 year ago
8

How much would the boiling point of water increase if mol of NaCl added 1 kg of water (Kb= 0.51 degrees Celsius /(mol/kg) for wa

ter i=2 for NaCl
Chemistry
1 answer:
Valentin [98]1 year ago
3 0

The boiling point of water would increase by 1°C if 1 mol of NaCl added 1 kg of water.

<h3>What would be the change in the boiling point of water when NaCl is added to it?</h3>

The addition of solute substances to solvent such as water results in a change in the boiling points and freezing points of the solvents.

The formula to calculate the change in temperature is given below:

ΔT = Kbmi

Where;

m is molality of the solution

Kb and i are constants.

Solving for change in temperature:

m = 1 mol/kg

ΔT = 0.51 * 1 * 2

ΔT = 1°C

The normal boiling point of water is 100°C

The new boiling point of the solution = (100 + 1) = 101 °C

In conclusion, addition of solute to solvent will result in change in the boiling point of the solvent.

Learn more about boiling point at: brainly.com/question/14557986

#SPJ1

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The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
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Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

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p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
2Li + H2SO4=Li2SO4 + H2 How many liters of hydrogen gas, H2 at STP can be produced from 3.0 moles of Li? The molar volume of a g
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Answer:

volume of H_2=33.6 litre

Explanation:

Firstly balance the given chemical equation,

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From the given balance equation it is clearly that,

2 mole of Li gives  1 mole of H2 gas

2 mole Li⇔1 mole H_2

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3 mole Li⇔1.5 mole H_2

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3 mole of Li will give 1.5 mole H2 gas

therefore volume of gas produced from 3 mole Li at STP = 1.5\times22.4 \frac{L}{mol}

volume of H2=33.6 litre

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Express 345.6 in scientific notation
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3.456 x 10^2 is the answer you're looking for.
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