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Question

In a 250 mL solution, 5 g of solute is dissolved. What is the concentration (mass by volume %) of the solution?

The correct answer is
2%

Calculating Mass by Volume Percentage Concentration

This question requires calculating the concentration of a solution in terms of mass by volume percentage. We are given the mass of the solute and the total volume of the solution.

Given Information

  • Mass of solute = 5 g
  • Volume of solution = 250 mL

Formula for Mass by Volume Percentage

The mass by volume percentage concentration is calculated using the following formula:

\( \text{Concentration (mass by volume } \%) = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100 \)

Step-by-Step Calculation

  1. Substitute the values: Plug the given mass of solute and volume of solution into the formula.

    \( \text{Concentration} = \frac{5 \text{ g}}{250 \text{ mL}} \times 100 \)

  2. Simplify the fraction: Divide the mass of solute by the volume of the solution.

    \( \frac{5}{250} = \frac{1}{50} \)

  3. Calculate the percentage: Multiply the result by 100.

    \( \text{Concentration} = \frac{1}{50} \times 100 = 2 \)

Result

The calculated concentration is 2% (mass by volume).

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Important Questions from Solutions

  1. Sugar is a _____ in a sugar solution.

    A. Solvent

    B. Solute

    C. Colloid

    D. Suspension

  2. When a solid body is partially or completely immersed in a fluid, the fluid exerts an upward force on the body. The magnitude of the force is equal to

    (1) the mass of the body

    (2) the weight of the displaced fluid by the body

  3. Calculate the molar mass of copper(II) sulfate pentahydrate, $CuSO_4 \cdot 5H_2O$.
    (Atomic masses: $Cu = 63.55 \text{ g/mol}$, $S = 32.07 \text{ g/mol}$, $O = 16.00 \text{ g/mol}$, $H = 1.01 \text{ g/mol}$)

  4. The pH value of 1 × 10 -8 (M) HCl is:

  5. The molar conductivity of 0.01 M acetic acid is 10 S cm2 mol−1. What is the dissociation constant of acetic acid? Choose the correct option.

    \(\left[\begin{array}{l}\Lambda_{\text{H}^{+}}^{\circ}=345 ~\text{S}~ \text{cm}^{2}\text{mol}^{-1} \\ \Lambda_{\text{CH}_{3}\text{COO}^{-}}^{\circ}=55~\text{S cm}^{2} \text{mol}^{-1}\end{array}\right]\)

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