QuestionJune 15, 2025

35.3 g of MgCl_(2) in the preparation of a 4.3 percent by mass solution. Round your answer to 2 significant digits. square g

35.3 g of MgCl_(2) in the preparation of a 4.3 percent by mass solution. Round your answer to 2 significant digits. square g
35.3 g of MgCl_(2) in the preparation of a 4.3 percent by mass solution. Round your answer to 2 significant digits. square g

Solution
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Answer

820 g Explanation 1. Define the percentage by mass formula The formula for percent by mass is **\(\text{Percent by mass} = \left(\frac{\text{mass of solute}}{\text{mass of solution}}\right) \times 100\)**. 2. Set up the equation Let ( x ) be the mass of the solution. Then, \(\frac{35.3}{x} \times 100 = 4.3\). 3. Solve for the mass of the solution Rearrange to find ( x ): \( x = \frac{35.3 \times 100}{4.3} \). 4. Calculate the mass of the solution \( x = \frac{3530}{4.3} \approx 820.9302 \). 5. Round to two significant digits The mass of the solution is approximately 820 g when rounded to two significant digits.

Explanation

1. Define the percentage by mass formula<br /> The formula for percent by mass is **\(\text{Percent by mass} = \left(\frac{\text{mass of solute}}{\text{mass of solution}}\right) \times 100\)**.<br /><br />2. Set up the equation<br /> Let ( x ) be the mass of the solution. Then, \(\frac{35.3}{x} \times 100 = 4.3\).<br /><br />3. Solve for the mass of the solution<br /> Rearrange to find ( x ): \( x = \frac{35.3 \times 100}{4.3} \).<br /><br />4. Calculate the mass of the solution<br /> \( x = \frac{3530}{4.3} \approx 820.9302 \).<br /><br />5. Round to two significant digits<br /> The mass of the solution is approximately 820 g when rounded to two significant digits.
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