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Temka [501]
1 year ago
14

What is the predicted change in the boiling point of water when 1.50 g of

Chemistry
1 answer:
dezoksy [38]1 year ago
7 0

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

\textsf {While} \:  \sf  {\Delta T_b}  \: \textsf{expression is used} \\  \textsf {for elevation of boiling point}

⠀

⠀

<u>The</u><u> </u><u>elevation</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>phenomenon</u><u> </u><u>in</u><u> </u><u>which</u><u> </u><u>there</u><u> </u><u>is</u><u> </u><u>increase</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>in</u><u> </u><u>solution</u><u>,</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>particular</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>is</u><u> </u><u>added</u><u> </u><u>to</u><u> </u><u>pure</u><u> </u><u>solvent</u><u>.</u>

⠀

⠀

\sf  \large \underline{The \:  formula \: to \:  be  \: used \:  in \:  this \:  question \:  is}  \\   \boxed{T_b = i \times  K_b \times  m}

⠀

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

⠀

<u>To</u><u> </u><u>find</u><u> </u><u>molality</u><u>,</u><u> </u><u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>divide</u><u> </u><u>no</u><u>.</u><u> </u><u>of</u><u> </u><u>moles</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>by</u><u> </u><u>weight</u><u> </u><u>of</u><u> </u><u>solution</u>

⠀

While first we need to no. of moles

\sf \implies no. \: of \: moles =  \frac{weight \: of \: solute}{molar \: mass \: of \: solute}  \\  \\ \implies \sf no. \: of \: moles =  \frac{1.5}{208.23}  \\  \\  \sf \implies  no. \: of \: moles = 0.0072

⠀

⠀

<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>

⠀

\sf  \hookrightarrow molality =  \frac{no.\: of \: moles}{weight \: of \: solution}  \\  \\  \sf  \hookrightarrow molality =  \frac{0.072}{1.5}  \\  \\  \sf  \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}

⠀

⠀

\textsf{ \large{ \underline{Now substituting the required values}}}

⠀

\sf \longmapsto \Delta T_b = 3  \times 0.51  \times 0.0048 \\  \\ \\     \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}

⠀

⠀

⠀

<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>

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b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

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Liquid EO                       -77.4

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CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

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A student notices that after two chemicals are mixed together the temperature of the mixture is higher than the temperature of t
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Now we will compare the moles of oxygen with Al.

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