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4 years ago

Calculate the percentage of a solution formed by dissolving 10g of glucose in 240g of water

1 answer:

8 0

X=240 g NaCl salt dissolves in solution.
Example: If we add 68 g sugar and 272 g water to 160 g solution having concentration 20 %, find final concentration of this solution.
Solution:
Mass of solution is 160 g before addition sugar and water.
100 g solution includes 20 g sugar
160 g solution includes X g sugar
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
X=32 g sugar
Mass of solute after addition=32 + 68=100 g sugar
Mass of solution after addition=272 +68 + 160=500 g
500 g solution includes 100 g sugar
100 g solution includes X g sugar
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
X= 20 % is concentration of final solution.

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What is the molarity of a solution that contains 224 grams of KOH in 2<br> liters of solution?

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Answer:

$\boxed {\boxed {\sf 2 \ M \ KOH}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

<h3>1. Grams to Moles </h3>

The first step is to convert the amount of grams given to moles. The molar mass is used. This found on the Periodic Table and it's the same value as the atomic mass, but the units are grams per mole.

We have 224 grams of KOH. Look up the molar masses for the individual elements.

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Since the compound's formula has no subscripts, 1 formula unit has 1 atom of each element. We can simply add the molar masses together to find KOH's molar mass.

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Use this number as a ratio.

$\frac {56.105 \ g\ KOH}{1 \ mol \ KOH}$

Multiply by the value we are converting: 224 g KOH

$224 \ g \ KOH *\frac {56.105 \ g\ KOH}{1 \ mol \ KOH}$

Flip the ratio so the units of grams KOH cancel.

$224 \ g \ KOH *\frac {1 \ mol \ KOH}{56.105 \ g\ KOH}$

$224 *\frac {1 \ mol \ KOH}{56.105}$

$\frac {224}{56.105} \ mol \ KOH$

$3.992514036 \ mol \ KOH$

<h3>2. Calculate Molarity </h3>

Remember molarity is moles per liter.

$molarity = \frac{moles}{liters}$

We just calculated the moles and we know there are 2 liters of solution.

$molarity = \frac{ 3.992514036 \ mol \ KOH}{ 2 \ L}$

$molarity= 1.996257018 \ mol \ KOH/ L$

<h3>3. Round and Convert Units </h3>

First, let's round. The original values have 3 and 1 significant figures. We go with the lowest number: 1. For the number we found, that is the ones place.

1.<u>9</u>96257018

The 9 in the tenths place tells us to round to 1 up to a 2

$2 \ mol \ KOH/ L$

Next, convert units. 1 mole per liter is equal to 1 molar or M.

$2 \ M \ KOH$

The molarity of the solution is <u>2 M KOH</u>

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The net ionic equation of a chemical reaction can be described as an equation that expresses only those ions, elements, or compounds, that directly contributed to that chemical reaction.

The chemical equation for the reaction of H₂CO₃ and LiOH:

H₂CO₃ (aq) + LiOH (aq) → Li₂CO₃ (aq) + 2H₂O (l)

The complete ionic equation for the above reaction H₂CO₃ and LiOH is:

2H⁺ (aq) + CO₃²⁻ (aq) + 2Li⁺ (aq) + 2OH⁻ → 2Li⁺ (aq) + CO₃²⁻(aq) + 2H₂O (l)

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