Question

A 6.30 g sample of a solid containing Ni is dissolved in 20.0 mL water. A 5.00 mL aliquot of this solution is diluted to 100.0 mL and analyzed in the lab. The analyzed solution was determined to contain 6.56 ppm Ni . 1. Determine the molar concentration of Ni in the 20.0 mL solution.2. Determine the mass in grams of Ni in the original sample
3. Determine the weight percent of Ni in the sample

Answer 1

Answer:

1. Molarity of  Ni in 20.0 mL water is 2.24 x 10⁻³ M

2. Amount of Ni in grams in the sample 2.63 x 10⁻³ g

3. Percentage of Ni in the sample 0.0417%

Explanation:

1. We can solve this part by using the dilution equation, then determining the molarity

The dilution equation:

$(C_{1})(V_{1})=(C_{2})(V_{2})$

Here, the final concentration (6.56 ppm), final volume (100.0 mL) and initial volume (5.00) is known (note that we are not using 20.0 mL in initial volume; as 5.00 mL was taken from it, so the concentration of 5 mL will be the same as of 20.0 mL).

$(C_{1})(5.00mL)=(6.56ppm)(100.0mL)\\ \\ C_{1}= 131 ppm$

For the molarity of Ni

$M_{Ni} = \frac{ppm (\frac{mg}{L}) }{molar mass (\frac{g}{moles}) .(1000\frac{mg}{g}) }\\ \\ M_{Ni} = \frac{131 (\frac{mg}{L})}{58.6934 (\frac{g}{moles}) .(1000\frac{mg}{g})}\\ \\ M_{Ni} =2.24X10^{-3}\frac{mol}{L}$

2. Mass in grams can be obtained from the Molarity of Ni in 20.0 mL of water

$Molarity=\frac{moles}{Volume (L)}\\ \\ Moles=(Volume (L))(Molarity)\\ \\ Moles=(0.02)(0.00224)\\ \\ Moles= 4.48X10^{-5} mol$

These moles can be used to determine the mass of Ni,

$n_{Ni}= \frac{mass}{molar mass}\\ \\ mass of Ni = (4.48X10^{-5})(58.6934)\\ \\ mass of Ni =2.63X10^{-3}$

3. Weight percent can be calculated by the mass determined in second part of the question

$Percent Wof Ni=(\frac{mass of Ni}{mass of sample}).100\\\\ Percent W of Ni=(\frac{2.63X10^{-3}}{6.30}).100\\ \\ Percent Wof Ni=0.0417$

The percentage of Ni in the sample is found to be 0.0417 %