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JulijaS [17]
3 years ago
10

What is the molarity of a solution that is made by dissolving 100.0 grams of lithium hydroxide in 750.0 mL of water?

Chemistry
1 answer:
vredina [299]3 years ago
3 0

Answer:

The answer to your question is: 5.6 M

Explanation:

Data

Molarity = ?

mass of LiOH = 100 g

final volume = 750 ml

MW LiOH = 7 + 16 + 1 = 24 g

Formula

Molarity = moles / volume

Process

                            24 g ------------------ 1 mol

                          100 g ------------------   x

                              x = (100 x 1) / 24

                              x = 4.2 moles of LiOH

Molarity = 4.2 / 0.750

Molarity = 5.6 M

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You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
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mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

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amount of X in 1g of Y = 0.4 x 1 /4.2

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for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

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amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

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formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

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