If 80 g of copper sulphate CuSO4.5H2O is dissolved in deionised water to make 5 L of solution. The c

Q: Why do we include the mass of water of crystallization ($5\text{H}_2\text{O}$) in the molar mass?

The water of crystallization is an integral part of the solid crystal lattice of copper sulphate pentahydrate. When you weigh 80 g of the salt, you are weighing the water molecules as well. Therefore, its molar mass must include the mass of these 5 water molecules.

Q: What if the volume was given in milliliters?

Molarity is strictly defined as moles of solute per liter of solution. If the volume were given in milliliters, you would first need to convert it to liters by dividing by 1000 before plugging it into the molarity formula.

Q: Why did we round off 64.11 to 64?

The question format typically expects an integer answer for such numerical value questions in JEE unless specified otherwise. 64.11 is closest to the integer 64.

Q: Does the addition of solute change the volume of the solution?

In this problem, it is stated that the solute is dissolved to *make* 5 L of solution. This means the final volume of the solution is exactly 5 L, so we don't need to worry about any volume changes upon dissolution.

Q: Can I use the formula $M_1V_1 = M_2V_2$ here?

No, the dilution formula $M_1V_1 = M_2V_2$ is used when you are adding solvent to an already existing solution to change its concentration. Here, we are preparing a fresh solution from a solid solute.

Question

If $80	ext{ g}$ of copper sulphate $	ext{CuSO}_4 \cdot 5	ext{H}_2	ext{O}$ is dissolved in deionised water to make $5	ext{ L}$ of solution. The concentration of the copper sulphate solution is $x 	imes 10^{-3}	ext{ mol L}^{-1}$. The value of $x$ is ......... .
[Atomic masses $	ext{Cu} = 63.54	ext{ u}$, $	ext{S} = 32	ext{ u}, 	ext{O} = 16	ext{ u}, 	ext{H} = 1	ext{ u}$]

Sigma Percentile · Accepted Answer

### The Art of Solution Making

Imagine you are in a chemistry lab, tasked with preparing a specific solution. You have a beaker, some deionised water, and exactly $80	ext{ g}$ of beautiful blue copper sulphate pentahydrate crystals, chemically known as $	ext{CuSO}_4 \cdot 5	ext{H}_2	ext{O}$. Your goal is to dissolve these crystals to make exactly $5	ext{ L}$ of solution.

But the real challenge isn't just making the solution; it's determining its exact concentration in a very specific format: $x 	imes 10^{-3}	ext{ mol L}^{-1}$. Let's break down the chemistry and the math to find the elusive value of $x$.

### Decoding the Molar Mass

Before we can talk about concentration, we need to speak the language of chemistry: moles. To convert our $80	ext{ g}$ of solid into moles, we first need the molar mass of $	ext{CuSO}_4 \cdot 5	ext{H}_2	ext{O}$.

This is where many students make a classic mistake—they forget the water of crystallization! The $5	ext{H}_2	ext{O}$ molecules are trapped within the crystal lattice, and their mass must be accounted for when you weigh the solid. Let's carefully add up the atomic masses:

$$M_w = 	ext{Cu} + 	ext{S} + 4 	imes 	ext{O} + 5 	imes (	ext{H}_2	ext{O})$$

Substituting the given atomic masses:

$$M_w = 63.54 + 32 + 4(16) + 5(18)$$

$$M_w = 63.54 + 32 + 64 + 90 = 249.54	ext{ g mol}^{-1}$$

### The Mole Concept in Action

Now that we have the molar mass, finding the number of moles ($n$) is a breeze. We simply divide the given mass by the molar mass:

$$n = \frac{80	ext{ g}}{249.54	ext{ g mol}^{-1}} \approx 0.3205	ext{ mol}$$

So, floating around in our $5	ext{ L}$ beaker are approximately $0.3205$ moles of copper sulphate.

### The Final Concentration

Molarity ($M$) is defined as the number of moles of solute per liter of solution. We have our moles, and we have our volume ($5	ext{ L}$). Let's plug them into the formula:

$$M = \frac{0.3205	ext{ mol}}{5	ext{ L}} = 0.0641	ext{ mol L}^{-1}$$

We have the concentration, but we aren't done yet. The question demands the answer in the format $x 	imes 10^{-3}$. Let's manipulate our decimal to match this format:

$$0.0641 = 64.1 	imes 10^{-3}$$

Comparing this to $x 	imes 10^{-3}$, we can clearly see that $x = 64.1$. Since these types of numerical questions typically require an integer answer, we round $64.1$ to the nearest integer, which gives us our final, triumphant answer: **64**.

Enter Numerical Value:

Visualized Solution

$Problem Setup$

$Molarity Formula$

$Molar Mass of CuSO_{4} \cdot 5 H_{2} O$

$Moles of Solute$

$Calculating Molarity$

$Formatting the Answer$

$Conclusion$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Art of Solution Making

Decoding the Molar Mass

The Mole Concept in Action

The Final Concentration

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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; } Problem 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 of CuSO4​⋅5H2​O

.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; } Formatting the Answer

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