Question

question 1
how many grams of mgco₃ are in 4.99 l of 0.0851 m mgco₃ solution?
(hint: once you have the moles of the solute find the grams)
0.00504 g
35.8 g
4.946 kg
1.44 g

Explanation:

Step1: Calculate moles of \( \text{MgCO}_3 \)

Molarity (\( M \)) is moles (\( n \)) per liter (\( V \)): \( M = \frac{n}{V} \). Rearranged, \( n = M \times V \).
Given \( M = 0.0851 \, \text{mol/L} \), \( V = 4.99 \, \text{L} \).
\( n = 0.0851 \, \text{mol/L} \times 4.99 \, \text{L} \approx 0.4246 \, \text{mol} \).

Step2: Find molar mass of \( \text{MgCO}_3 \)

Molar mass of \( \text{Mg} = 24.305 \, \text{g/mol} \), \( \text{C} = 12.011 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass (\( MM \)): \( 24.305 + 12.011 + 3(16.00) = 84.316 \, \text{g/mol} \).

Step3: Calculate mass from moles

Mass (\( m \)) = \( n \times MM \).
\( m = 0.4246 \, \text{mol} \times 84.316 \, \text{g/mol} \approx 35.8 \, \text{g} \)? Wait, no—wait, recalculate \( n \): \( 0.0851 \times 4.99 = 0.0851 \times 5 - 0.0851 \times 0.01 = 0.4255 - 0.000851 = 0.4246 \, \text{mol} \). Then \( 0.4246 \times 84.316 \approx 35.8 \)? But wait, the options have 35.8 g. Wait, but let's check again. Wait, maybe I miscalculated. Wait, \( 0.0851 \times 4.99 = 0.0851 \times 4.99 \). Let's compute 0.0851 4 = 0.3404, 0.0851 0.99 = 0.084249, total = 0.3404 + 0.084249 = 0.424649 mol. Then molar mass of MgCO3: Mg=24.305, C=12.01, O=163=48. So 24.305+12.01+48=84.315 g/mol. Then 0.424649 84.315 ≈ 0.4246 84.315. Let's compute 0.4 84.315 = 33.726, 0.0246 * 84.315 ≈ 2.074, total ≈ 35.8 g. So the answer is 35.8 g.

Answer:

35.8 g (corresponding to the option "35.8 g")