题目
题目

M2LA F25 Expt 2 PreLab Quiz

多重下拉选择题

Before coming to lab a plan is needed for the creation of 5 standard aqueous solutions of (NH4)3Fe(ox)3.3H2O with concentrations between 1.0 x 10-3 M and 7.5 x 10-3 M.   The first step is to create 25 mL of 0.012 M stock solution with (NH4)3Fe(ox)3.3H2O(s) and deionized water.  What is the formula weight of (NH4)3Fe(ox)3.3H2O? [ Select ] 392,03 406.02 428.06 391.99 What mass (in grams) of (NH4)3Fe(ox)3.3H2O is required to make such a solution? [ Select ] 0.040 0.121 1.34 0.128 What piece of glassware holds the 25 mL? [ Select ] graduated cylinder Erlenmeyer flask beaker volumetric flask If the digital scale in lab reports values to the thousandths place, how many significant figures can the actual stock solution concentration be reported with? [ Select ] 3 2 4 1 5  

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思路分析
This problem consists of several linked parts about preparing a stock solution and choosing appropriate equipment and significant figures. I will assess each option in turn and explain why it is or isn’t appropriate for the stated task. First sub-question: Determine the formula weight of (NH4)3Fe(ox)3·3H2O. - Option 428.06: To evaluate, sum the atomic weights of all atoms in the formula unit. Ammonium (NH4)3 contributes 3 × (14.01 + 4×1.008) = 3 × (14.01 + 4.032) ≈ 3 × 18.042 ≈ 54.126 g/mol. Iron in the complex contributes about 55.845 g/mol. The oxalate (ox)3 units are each C2O4 with a charge balance, but the exact molecular weight would require summing 3 × (C2O4) contributions and the 3 waters of hydration (3 × 18.015 g/mol). The final total is typically around the 428 g/mol region, so 428.06 g/mol is plausible. If you compare to the other numbers, 406.02, 392.03, and 391.99, those deviate by enough to signal they are unlikely correct given the hydrated complex’s mass; thus 428.06......Login to view full explanation

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