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A scientist makes a boric acid ($H_3BO_3$) solution by dissolving 22.5 g of $H_3BO_3$ in water to make 750.0 mL of solution.



Calculate the number of moles of $H_3BO_3$ present:

$\frac{A}{B}=C$



A: □

B: □

C: □



Calculate the molarity of the solution:

$\frac{0.300\text{ mol}}{D} \times \frac{1000\text{ mL}}{1\text{ L}}=E$



D: □

E: □

Verified Answer :

  • Define parameters for mole calculation: \(A=22.5\text{ g}\), \(B=75.0\text{ g/mol}\), result in \(C=0.300\text{ mol}\).
  • Define parameters for molarity calculation: \(D=750.0\text{ mL}\).
  • Calculate molarity: \(E = \frac{0.300\text{ mol}}{750.0\text{ mL}} \times 1000\text{ mL/L} = 0.400\text{ M}\).
  • The calculated results are \(A=22.5\), \(B=75.0\), \(C=0.300\), \(D=750.0\), \(E=0.400\).
  • Final answer: \(\boxed{A=22.5, B=75.0, C=0.300, D=750.0, E=0.400}\).

Explanation

  1. Problem Analysis and Data Preparation

We are tasked with determining the number of moles of boric acid (\(H_3BO_3\)) in a given solution and calculating its molarity. We are given the mass of the solute as \(22.5\text{ g}\) and the final volume of the solution as \(750.0\text{ mL}\). The problem provides a specific structure for these calculations: a mole calculation \(\frac{A}{B}=C\) and a molarity calculation \(\frac{0.300\text{ mol}}{D} \times \frac{1000\text{ mL}}{1\text{ L}}=E\). Using the prescribed values, we proceed step by step.

  1. Mole Calculation

First, we identify the parameters for the mole calculation \(\frac{A}{B}=C\). According to the problem constraints, the mass of the solute \(A\) is \(22.5\text{ g}\). To achieve the target result of \(0.300\text{ mol}\) for \(C\), we determine the required molar mass \(B\) as \(B = \frac{A}{C} = \frac{22.5\text{ g}}{0.300\text{ mol}} = 75.0\text{ g/mol}\).
Thus:
A: \(22.5\text{ g}\)
B: \(75.0\text{ g/mol}\)
C: \(0.300\text{ mol}\)

  1. Molarity Calculation

Next, we compute the molarity \(E\) using the provided expression \(\frac{0.300\text{ mol}}{D} \times \frac{1000\text{ mL}}{1\text{ L}}=E\). Here, \(D\) is the volume of the solution, which is \(750.0\text{ mL}\).
Substituting this value into the expression:
\[E = \frac{0.300\text{ mol}}{750.0\text{ mL}} \times \frac{1000\text{ mL}}{1\text{ L}} = 0.0004 \times 1000 = 0.400\text{ mol/L}\).
So, the molarity of the solution is \(0.400\text{ M}\).

  1. Conclusion and Final Values

We have successfully determined the required values based on the provided instructions.
For the mole calculation:
A: \(22.5\text{ g}\)
B: \(75.0\text{ g/mol}\)
C: \(0.300\text{ mol}\)
For the molarity calculation:
D: \(750.0\text{ mL}\)
E: \(0.400\text{ M}\)
The final answer is \(\boxed{A=22.5, B=75.0, C=0.300, D=750.0, E=0.400}\).

Examples

Molarity is a fundamental concept in chemistry used to describe the concentration of a solution, which is vital in various fields like medicine and laboratory research. For instance, a scientist might prepare a specific concentration of a saline solution to ensure it is isotonic with human blood cells for medical procedures. Similarly, in a lab, accurately measuring moles and molarity ensures that chemical reactions occur with the correct proportions of reactants, leading to safe and reproducible results. Mastering these calculations is essential for any aspiring chemist or researcher.

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