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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<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>
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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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<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>
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While first we need to no. of moles
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<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>
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<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>
Specialized proteins which function as catalysts for organic reactions are enzymes.
Alkali are soluble bases, however not all bases are soluble in water, therefore not all bases are Alkali.
Answer:
Chemists investigate the properties of matter at the level of atoms and molecules. They measure proportions and reaction rates in order to understand unfamiliar substances and how they behave, or to create new compounds for use in a variety of practical applications.
Explanation: