A solution was made by combining 145 (±1) mL of 0.250 (±0.003) M NaCl with 275.0

Question

A solution was made by combining 145 (±1) mL of 0.250 (±0.003) M NaCl with 275.0 (±0.5) mL of 6.00 (±0.01) wt% NaCl (assume this solution’s density is 1.0413 g/mL and the molar mass of NaCl is 58.443 g/mol, both with negligible error) in a 500.00 mL volumetric flask, and filling with water up to the line.

How many mmol of NaCl are in the final solution? Include absolute error in your number of millimoles and round your answer to an appropriate number of sig figs.

(HINT: First find the number of millimoles of NaCl, ignoring the error. Then go back and propagate error properly through each calculation step. )

Explanation / Answer

First calcualte the number of moels in solution as follows:

145 (±1) mL of 0.250 (±0.003) M NaCl

Number of moles = volume * molarity

= 0.145 L * 0.250 M or moles / L

=0.03625 mol

275.0 (±0.5) mL of 6.00 (±0.01) wt% NaCl

density =1.0413 g/mL and the molar mass of NaCl = 58.443 g/mol,

Use the density to calculate the weight of 1 liter of solution-
1000 mL x 1.0413 g/mL = 1041.3 grams

Use the percent to calculate the mass of NaCl in 1041.3 g (1.000 liters)-
1,041.3 x 6 % = 62.478 g NaCl

Now you know the weight of HCl per liter. Convert the weight of NaCl to moles of HCl-
62.478 g / 58.443 g/mol = 1.069 moles of NaCl

Total number of moles = 0.03625 mol + 1.069 moles of NaCl
=1.1053 mol = 1105.3 mmol

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