Question

The principal investigator (PI) for the laboratory you are doing research in has tasked with you with preparing 250 mL of an aqueous zinc sulfate (ZnSO4) solution with a concentration of approximately 50. mM- it is not critical for the solution to be exactly 50 mM, it is just critical to know what it exactly is. Proceed through the following steps to quantify the concentration and uncertainty for the solution you prepare. For this question use the following molar mass values: Zn = 65.38(2), S-32.068(9), O= 15.9994(4) g/mol. a. Consider the uncertainties associated with the average atomic molar masses of Zn, S, and O to calculate and report the confidence interval for the molar mass of ZnSO4. When you report this, go ahead and report he molar mass with 5 significant digits plus one extra digit beyond significance and report the uncertainty associated with the molar mass with 1 significant digit plus one extra digit beyond significance (e.g. MM333.33s t 0.033 g/mol) b. What is the percent relative uncertainty in the molar mass of ZnSO4? Report this quantity rounded to 1 digit. c. You use an analytical balance to mass out 2.0184(4)g of ZnSO4, add it to an uncalibrated 250.0(2) mL volumetric flask, and carefully fill it with nanopure water exactly to the fill line. Considering the uncertainty in the molar mass, the balance, and the volumetric flask - report the confidence interval for the molarity - in units of millimolarity, mM - of the solution you have prepared. When presenting your final confidence interval, round your absolute uncertainty to 1 digits and report the mean concentration with 4 significant digits

Answer 1

Answer A)

The molar mass is simply the sum of atomic mass with coefficient .Hence the molar mass of ZnSO4 is sum of atomic masses of element Zn S O .

Molar mass = 65.38+32.068+((15.9994)×4)

Molar mass = 161.445 (+-)0.045 g / mole.

Answer B)

Percent Relative uncertainty or error is calculated as :

Formula = Absolute error / measured value. × 100

Then,absolute error is 0.045i.e. 0.05 and measured value is 161.44

Then (0.05/161.44)×100

% RE = 3.0 %

Answer C)

The given quantity is in grams convert in to milligram i.e.2.0184 = 2018 mm .

Hence we know that ,

X = CMV/ 1000

Where X = amount of substance ,C = con.of solution ,M = molar mass ,V = volume to be prepared.

Then, 2018.4 = (C × 161.445 × 250 )/ (1000)

C = 50.00 millimolar .