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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![\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}](https://tex.z-dn.net/?f=%20%5Ctextsf%20%7BWhile%7D%20%5C%3A%20%20%5Csf%20%20%7B%5CDelta%20T_b%7D%20%20%5C%3A%20%5Ctextsf%7Bexpression%20is%20used%7D%20%5C%5C%20%20%5Ctextsf%20%7Bfor%20elevation%20of%20boiling%20point%7D)
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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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![\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}](https://tex.z-dn.net/?f=%20%5Csf%20%20%5Clarge%20%5Cunderline%7BThe%20%5C%3A%20%20formula%20%5C%3A%20to%20%5C%3A%20%20be%20%20%5C%3A%20used%20%5C%3A%20%20in%20%5C%3A%20%20this%20%5C%3A%20%20question%20%5C%3A%20%20is%7D%20%20%5C%5C%20%20%20%5Cboxed%7BT_b%20%3D%20i%20%5Ctimes%20%20K_b%20%5Ctimes%20%20m%7D)
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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
![\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](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7Bweight%20%5C%3A%20of%20%5C%3A%20solute%7D%7Bmolar%20%5C%3A%20mass%20%5C%3A%20of%20%5C%3A%20solute%7D%20%20%5C%5C%20%20%5C%5C%20%5Cimplies%20%5Csf%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7B1.5%7D%7B208.23%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%5Cimplies%20%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%200.0072%20)
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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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![\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}](https://tex.z-dn.net/?f=%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%20%20%5Cfrac%7Bno.%5C%3A%20of%20%5C%3A%20moles%7D%7Bweight%20%5C%3A%20of%20%5C%3A%20solution%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%20%20%5Cfrac%7B0.072%7D%7B1.5%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%200.048%20%5C%3A%20mol%20%7Bkg%7D%5E%7B%20-%201%7D%20)
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![\textsf{ \large{ \underline{Now substituting the required values}}}](https://tex.z-dn.net/?f=%20%5Ctextsf%7B%20%5Clarge%7B%20%5Cunderline%7BNow%20substituting%20the%20required%20values%7D%7D%7D%20)
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![\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%5CDelta%20T_b%20%3D%203%20%20%5Ctimes%200.51%20%20%5Ctimes%200.0048%20%5C%5C%20%20%5C%5C%20%5C%5C%20%20%20%20%20%5Cboxed%7B%20%5Ctt%7B%20%5Clongmapsto%20%5CDelta%20T_b%20%3D0.00735%7B%20%5Cdegree%7DC%7D%7D)
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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>
The answer is 40.
We can solve this by finding out the number of protons, and neutrons. Atomic number of an element means the number of protons in that element. So, the atom has 30 protons if the atomic number is 30.
On the other hand, mass number is the total number of protons and neutrons, but not electrons, because they're too light comparing to the other 2. Therefore, we can simply solve the number of neutrons in the atom by subtracting the number of protons from the mass number. 70 - 30 = 40.
Therfore, the number of neutrons is 40.
Electrons in an atom are found on different electron shells depending on how much energy they possess, and they determine how atoms will interact with each other. The outermost electron shell holds the valance electrons.