The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4 . 2H2O) in 250 mL o

Q: Why do we include the mass of water in the molar mass of oxalic acid?

The chemical formula given is $H_2C_2O_4 \cdot 2H_2O$, which means it is a hydrated crystal. When you weigh 6.3 g of this substance, the water molecules are part of that mass. If you ignore them, your mole calculation will be incorrect.

Q: Can I assume the volume of the solution is exactly 250 mL?

Yes. In dilute solutions where the density is not provided, it is standard practice to assume the volume of the solvent (water) is approximately equal to the volume of the final solution.

Q: How do I quickly calculate the moles without a calculator?

Notice the numbers: $6.3$ and $126$. If you multiply $6.3$ by $2$, you get $12.6$. So, $6.3 / 126$ is the same as $12.6 / (2 \times 126) = 0.1 / 2 = 0.05$.

Q: What if the question asked for molality instead of molarity?

For molality, you would need the mass of the solvent in kg. Since the density of water is $1 \text{ g/mL}$, 250 mL of water is 250 g or 0.25 kg. The molality would be $0.05 \text{ mol} / 0.25 \text{ kg} = 0.2 \text{ m}$.

Q: Why did we convert $0.2$ to $20 \times 10^{-2}$?

The question specifically asks for the value of $x$ when the molarity is expressed as $x \times 10^{-2}$. To find $x$, we must format our answer to match this exact scientific notation.

Question

The molarity of the solution prepared by dissolving $6.3 	ext{ g}$ of oxalic acid ($H_2C_2O_4 \cdot 2H_2O$) in $250 	ext{ mL}$ of water in $	ext{mol L}^{-1}$ is $x 	imes 10^{-2}$. The value of $x$ is ......... (Nearest integer)
[Atomic mass H = 1.0, C = 12.0, O = 16.0]

Sigma Percentile · Accepted Answer

The problem asks us to find the molarity of a solution prepared by dissolving $6.3 	ext{ g}$ of oxalic acid dihydrate ($H_2C_2O_4 \cdot 2H_2O$) in $250 	ext{ mL}$ of water.

### Analyzing the Setup

Imagine you are in a chemistry lab. You have a beaker containing $250 	ext{ mL}$ of water. Now, you carefully weigh out $6.3 	ext{ g}$ of oxalic acid dihydrate and add it to the beaker. Our goal is to find the molarity of this resulting solution.

To find the molarity, we need to recall its definition. Molarity is simply the number of moles of solute dissolved per liter of the solution. So, we need two things: the moles of oxalic acid, and the volume of the solution in liters.

### The Master Equation

First, let's calculate the molar mass of our solute, oxalic acid dihydrate. The formula is $H_2C_2O_4 \cdot 2H_2O$. Don't forget the water of crystallization! We add the atomic masses: $2$ for hydrogen, $24$ for carbon, $64$ for oxygen, and $36$ for the two water molecules.

$$M_w = 2(1) + 2(12) + 4(16) + 2(18) = 126 	ext{ g mol}^{-1}$$

Now that we have the molar mass, finding the number of moles is a piece of cake. We just divide the given mass, $6.3 	ext{ g}$, by the molar mass, $126 	ext{ g mol}^{-1}$. $6.3$ is exactly half of $12.6$, so this simplifies beautifully to $0.05 	ext{ moles}$.

$$n = \frac{6.3}{126} = 0.05 	ext{ mol}$$

### Final Calculation

Let's plug these values into our molarity formula. The moles are $0.05$. The volume is $250 	ext{ mL}$, which we must convert to liters by dividing by $1000$. So, we get $0.05$ divided by $0.25$. This gives us a molarity of $0.2 	ext{ mol L}^{-1}$.

$$M = \frac{0.05 	ext{ mol}}{0.25 	ext{ L}} = 0.2 	ext{ mol L}^{-1}$$

The question states that the molarity is $x 	imes 10^{-2}$. So, we equate our result, $0.2$, to this expression. To match the powers of ten, we can write $0.2$ as $20 	imes 10^{-2}$.

$$0.2 = 20 	imes 10^{-2} = x 	imes 10^{-2}$$

Comparing both sides, we find that $x$ is exactly **$20$**. And that's our final answer!

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$M o l a r i t y F or m u l a$

$M o l a r M a ss C a l c u l a t i o n$

$M o l es o f S o l u t e$

$C a l c u l a t in g M o l a r i t y$

$F in d in g x$

$C o mm o n P i t f a l l$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molarity Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molar Mass Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Solute

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Molarity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Common Pitfall