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eimsori [14]
3 years ago
8

The mole fraction of iodine, i2, dissolved in dichloromethane, ch2cl2, is 0.115. what is the molal concentration, m, of iodine i

n this solution?
Chemistry
1 answer:
german3 years ago
5 0
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:

0.115 mol I₂

1 - 0.115 = 0.885 mol CH₂Cl₂

We need moles of solute, which we have, and must convert our moles of solvent to kg:

0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂

We can now calculate the molality:

m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂

The molality of the iodine solution is 1.53.
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a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula
kolezko [41]

The chemical compound's empirical formula is NS.

The chemical compound's molecular formula is N4S4.

<h3>What does a chemical empirical formula look like?</h3>
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<h3>What sort of empirical formula would that be?</h3>
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learn more about empirical formula here

brainly.com/question/1603500

#SPJ4

the question you are looking for is

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?

3 0
2 years ago
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
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