Question

How many grams of solute should be added in \(100 \mathrm{~g}\) water to get a solution of density \(1.2 \mathrm{~g} / \mathrm{m} 1\) and strength \(5 \%\) (w/v)? (a) \(5 \mathrm{~g}\) (b) \(6 \mathrm{~g}\) (c) \(4.17 \mathrm{~g}\) (d) \(4.35 \mathrm{~g}\)

Short answer

a) 5g b) 6g c) 4.17g d) 4.35g Answer: None of the given options lead to the required density of 1.2 g/mL and strength of 5% (w/v). The question may contain errors or insufficient information.

Step-by-step solution

Calculate the volume of the solution using the given density

To calculate the volume, we will use the formula: density = mass / volume Where the density is given as 1.2 g/mL, and we have a total mass of 100g of water + grams of solute. We will call the grams of solute "x". 1.2 g/mL = (100g + x) / volume Since the solute and solvent are being mixed, we can assume that the volume is additive. That means that the volume of the solution equals the sum of the volume of the solute and the volume of the solvent (water). We know that 100g of water has a volume of 100mL because water has a density of 1 g/mL. volume = 100 mL + volume of solute Now, we can substitute this expression for the volume into the density equation. 1.2 g/mL = (100g + x) / (100 mL + volume of solute)

Calculate the volume of the solute using the given strength

We can use the strength percentage given (5% w/v) to find the quantity of solute in 100mL of the solution. 5% w/v = (grams of solute) / (100 mL) Re-arranging the formula to solve for grams of solute: grams of solute = 0.05 * 100 mL grams of solute = 5g Since the solute has a mass of 5g, we can re-write the expression for the volume of the solute: volume of solute = 5g / density of solute

Solve for the grams of solute (x)

Now that we have expressions for both the volume and grams of solute, we can substitute them back into the original density equation: 1.2 g/mL = (100g + x) / (100 mL + 5g / density of solute) As the density of the solute cannot be computed in this case, we have to find the value of x with the information given. On analysis we find out that we have to solve for x, taking into account the options given in the question. We will try each of the options given to see which one gives us a density of 1.2 g/mL (a) x = 5g 1.2 g/mL = (100g + 5g) / (100 mL) 1.2 g/mL = 105g / 100 mL 1.2 g/mL = 1.05 g/mL (This does not match the required density) (b) x = 6g 1.2 g/mL = (100g + 6g) / (100 mL) 1.2 g/mL = 106g / 100 mL 1.2 g/mL = 1.06 g/mL (This does not match the required density) (c) x = 4.17g 1.2 g/mL = (100g + 4.17g) / (100 mL) 1.2 g/mL = 104.17g / 100 mL 1.2 g/mL ≈ 1.04 g/mL (This does not match the required density) (d) x = 4.35g 1.2 g/mL = (100g + 4.35g) / (100 mL) 1.2 g/mL = 104.35g / 100 mL 1.2 g/mL ≈ 1.04 g/mL (This does not match the required density) From the above calculations, we can see that none of the options give us the required density of 1.2 g/mL. This indicates that there might be an error in the question, or that the question might contain insufficient information.

Key concepts

Understanding Solution Density

In chemistry, solution density refers to the mass per unit volume of a solution. For our given problem, the density is stated to be 1.2 g/mL, which means that, ideally, every milliliter of the solution should weigh 1.2 grams.
This concept is crucial because it helps to identify how concentrated or dilute a solution is based on how heavy it is compared to its volume.
When given a problem involving solution density, pay attention to whether mass and volume changes are additive after mixing solute and solvent. This helps decide if the density formula can easily apply.
Remember, with density, think of it as an indicator of how packed molecules are in a solution. The higher the density, the more tightly packed the molecules are.

Calculating the Grams of Solute

Finding the grams of solute is about determining how much of a substance (solute) is present in a solution. The problem specifies a strength of 5% w/v, meaning 5 grams of solute should dissolve in every 100 mL of solution.
The goal is to calculate how many grams need to be added to the solvent (water) to achieve this percentage.
You can calculate this by rearranging the given strength formula:

• Strength (w/v) = (grams of solute) / (volume of solution in mL)
• grams of solute = Strength * volume of solution

For 5% w/v strength in 100 mL, you have:

• grams of solute = 0.05 * 100 = 5 grams

This calculation tells us that initially, you should theoretically need 5 grams of solute, but be aware actual requirements can be affected by factors like incomplete solubility or volume expansion.

Volume Calculation After Mixing

Volume calculation in solutions involves the sum of the volumes of both the solvent and the solute after mixing. However, it isn't always straightforward due to interactions between solute and solvent.
In our problem, water is the solvent with a known density (1 g/mL), which allows us to simplify volume calculation as the same number of grams equals the same number of mL.
After adding the solute, the total volume must equal the sum of the solvent's volume and the solute's volume (parts-per-volume may not strictly apply in real-world cases, but it’s a good starting point).
Ensuring total mass matches the stated density (1.2 g/mL) while considering total volume (volume of water plus volume cordoned by solute) should guide your calculations.

Mixing Solute and Solvent

Mixing solute and solvent is more than just adding one to the other. It’s about understanding how they interact to form a uniform solution. The process can affect the final density and volume.
Start by recognizing the solute's role, which dissolves into the solvent, dispersing its particles among the solvent's molecules.
Key factors influencing this process include:

• Temperature: Often impacts how much solute can dissolve.
• Agitation: Stirring can improve dissolution rate.

After mixing, observing properties such as clarity, presence of residue, and actual volume can help ensure the solution meets theoretical predictions. Mixing results in a balanced combination of solute and solvent, reflecting the sought concentration and density.