Answer:
0.00735°C
Explanation:
By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water
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[tex] \textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}[/tex]
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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.
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[tex] \sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}[/tex]
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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)
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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)
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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.
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To find molality, we have to divide no. of moles of solute by weight of solution
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While first we need to no. of moles
[tex] \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 [/tex]
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Now, we will find molality
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[tex] \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} [/tex]
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[tex] \textsf{ \large{ \underline{Now substituting the required values}}} [/tex]
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[tex] \sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}[/tex]
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Henceforth, the change in boiling point is 0.00735°C.
