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Sever21 [200]
3 years ago
10

A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th

e molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl
Chemistry
1 answer:
Kay [80]3 years ago
7 0

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

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0.00735°C

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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}

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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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\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}

⠀

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

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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}}

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