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A 5.25% solution of a substance is isoto...

A `5.25%` solution of a substance is isotonic with a `1.5%` solution of urea (molar mass `= 60g mol^(-1)`) in the same solvent. If the densities of both the solutions are assumed to be equal to `1.0 g cm^(-3)`, molar mass of the substance will be:

A

`90.0g mol^(-1)`

B

`115.0g mol^(-1)`

C

`105.0g mol^(-1)`

D

`210.0g mol^(-1)`

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To solve the problem of finding the molar mass of a substance that is isotonic with a urea solution, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Concept of Isotonic Solutions**: - Two solutions are isotonic when they have the same osmotic pressure. This means that the concentration of solute particles in both solutions is the same. 2. **Identify Given Data**: - The concentration of the substance solution is `5.25%`. - The concentration of the urea solution is `1.5%`. - The molar mass of urea is `60 g/mol`. - The density of both solutions is assumed to be `1.0 g/cm³`. 3. **Convert Percentage Concentrations to Mass**: - For the `5.25%` solution: - This means `5.25 g` of solute in `100 mL` of solution. - For the `1.5%` urea solution: - This means `1.5 g` of urea in `100 mL` of solution. 4. **Calculate the Molarity of Each Solution**: - Molarity (C) is given by the formula: \[ C = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \] - For the urea solution: \[ C_{\text{urea}} = \frac{1.5 \text{ g}}{60 \text{ g/mol} \times 0.1 \text{ L}} = \frac{1.5}{6} = 0.25 \text{ mol/L} \] 5. **Set Up the Equation for the Substance**: - Since the solutions are isotonic, we can set the molarity of the substance equal to that of urea: \[ C_{\text{substance}} = C_{\text{urea}} = 0.25 \text{ mol/L} \] - The molarity of the substance can also be expressed as: \[ C_{\text{substance}} = \frac{5.25 \text{ g}}{M \times 0.1 \text{ L}} \] - Where \( M \) is the molar mass of the substance. 6. **Equate the Two Molarities**: \[ \frac{5.25 \text{ g}}{M \times 0.1 \text{ L}} = 0.25 \text{ mol/L} \] 7. **Solve for Molar Mass (M)**: - Rearranging the equation gives: \[ 5.25 = 0.25 \times M \times 0.1 \] \[ 5.25 = 0.025M \] \[ M = \frac{5.25}{0.025} = 210 \text{ g/mol} \] 8. **Conclusion**: - The molar mass of the substance is `210 g/mol`.

To solve the problem of finding the molar mass of a substance that is isotonic with a urea solution, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Concept of Isotonic Solutions**: - Two solutions are isotonic when they have the same osmotic pressure. This means that the concentration of solute particles in both solutions is the same. 2. **Identify Given Data**: ...
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