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3 years ago

what is the boiling point of the solution resulted from the dissolving of 32.5g of NaCl in 250.0g of water?

Chemistry

1 answer:

suter [353]3 years ago

8 0

Answer:

The boiling point of this solution is 102.28 °C

Explanation:

Step 1: Data given

Mass of Nacl = 32.5 grams

Molar mass of NaCl = 58.45 g/mol

Mass of Water = 250 grams

Boiling point of water = 100°C

Step 2: Calculate number of moles

Number of moles = mass of NaCl / Molar mass of NaCl

Number of moles = 32.5 grams /58.45 g/mol = 0.556 moles

Step 3: Calculate molality

Molality = Number of moles / mass of water

Molality = 0.556 moles / 0.250 kg of water

Molality = 2.224 molal

NaCl releases twice as many moles of ions, the total molality of this solution is also twice: 4.448 molal

Step 4: Calculate boiling point

dT =( 0.512 C / molal)*4.448 molal)

dT = 2.28

The boiling point of this solution is 100 °C + 2.28 °C = 102.28 °C

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horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

Ni 12.7 g

CO 1.98 g

$Ni(CO)_4$ 0.597 g

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Answer: The value of equilibrium constant for the reaction is 2448.1

Explanation:

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of $Ni(CO)_4$ = 0.597 g

Volume of container = 2.6 L

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

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

$\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M$

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For the given chemical reaction:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

The expression of equilibrium constant for the reaction:

$K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}$

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1$

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