Please help! I just need help determining the third part! 4.8841 g of Na2CrO4 (M

Question

Please help! I just need help determining the third part!

4.8841 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assuming the solution has a density of 1.00 g/mL, what is the concentration of Na (MW = 22.9898 g/mol) in the solution in units of a) molarity (M)?

MY ANSWER:

irst the number of moles of Na2CrO4 = 4.8841 / 161.97 = 0.030 mol

hence number of moles of Na = 2 * 0.03 = 0.06 mol

now concentration of Na = number moles of na / volume in L

= [0.06 / 100 ] * 1000 = 0.6 M

hence the molarity of Na = 0.6 M

b) parts per thousand (ppt)?

MY ANSWER: mass of 0.6 M Na

molarity = given mass /molar mass * volume

0.6 = X / 22.9898 * 0.100

X = 1.379 gram

Now mass of solution

given that 1 .0 density hence mass of solution is 100 gram

Hence we get 1.379 / 100 = 13.79 ppt

hence the ppt is 13.79

c) 10.0 mL of the solution is then diluted to a final volume of 250.0 mL. What is the concentration of Na in the diluted solution in units of parts per million (ppm)?

PLEASE NEED HELP WITH PART C

Explanation / Answer

M1V1= M2V2

10*0.6 = M2 * 250 M2 is the concentration of Na after dilution

M2 = 6 / 250

M2 = 0.024 g/ L

M2 = 24X 10-6 ppm

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