# Answer to Question #29881 in Inorganic Chemistry for Wael Elbiltagi

Question #29881

1st Question:-

What is the contraction of the Na and K if we take 1ml from Na (10ppm) and 1ml from K (10ppm) and dissolved them together with dist. water to be 10 ml mixed soln.?

2nd Question:-

What is the contraction of the Na and K if we take 1ml from a mixture contains Na (10ppm) and K(10ppm) and dissolved it with dist. water to be 10 ml mixed soln.?

What is the contraction of the Na and K if we take 1ml from Na (10ppm) and 1ml from K (10ppm) and dissolved them together with dist. water to be 10 ml mixed soln.?

2nd Question:-

What is the contraction of the Na and K if we take 1ml from a mixture contains Na (10ppm) and K(10ppm) and dissolved it with dist. water to be 10 ml mixed soln.?

Expert's answer

1) 1mL = 0.01 mg Na, 10 ppm Nа

1 mL = 0.01 mg К, 10 ppm К

To calculate the concentration we can use the formula

C(Na) = n(Na) / V solution

C(K) = n(K) / V solution

To calculate the amount of substance of Na and K we can use the formula

n(Na) = m/M

n(K) = m/M

n(Na) = 0.0001 g / 23 g/mol = 4.35e-6 mol

n(K) = 0.0001 g / 39 g/mol = 2.56e-6 mol

So the concentration will be:

C(Na) = 4.35e-6 mol / 0.01 L = 0.000435 mol/L

C(K) = 2.56e-6 mol / 0.01 L = 0.000265 mol/L

Answer: 0.000435 mol/L, 0.000265 mol/L

2) 1 mL = 0.005 mg Na + 0.005 mg K

To calculate the concentration we can use the formula

C(Na) = n(Na) / V solution

C(K) = n(K) / V solutionto calculate the amount of substance of Na and K we can use the formulan(Na) = m/M

n(K) = m/M

n(Na) = 0.00005 g / 23 g/mol = 2.17e-6 mol

n(K) = 0.00005 g / 39 g/mol = 1.28e-6 mol

So the concentration will be:

C(Na) = 2.17e-6 mol / 0.01 L = 0.000217 mol/L

C(K) = 1.28e-6 mol / 0.01 L = 0.000128mol//L

Answer: 0.000217mol/L, 0.000128mol/L

1 mL = 0.01 mg К, 10 ppm К

To calculate the concentration we can use the formula

C(Na) = n(Na) / V solution

C(K) = n(K) / V solution

To calculate the amount of substance of Na and K we can use the formula

n(Na) = m/M

n(K) = m/M

n(Na) = 0.0001 g / 23 g/mol = 4.35e-6 mol

n(K) = 0.0001 g / 39 g/mol = 2.56e-6 mol

So the concentration will be:

C(Na) = 4.35e-6 mol / 0.01 L = 0.000435 mol/L

C(K) = 2.56e-6 mol / 0.01 L = 0.000265 mol/L

Answer: 0.000435 mol/L, 0.000265 mol/L

2) 1 mL = 0.005 mg Na + 0.005 mg K

To calculate the concentration we can use the formula

C(Na) = n(Na) / V solution

C(K) = n(K) / V solutionto calculate the amount of substance of Na and K we can use the formulan(Na) = m/M

n(K) = m/M

n(Na) = 0.00005 g / 23 g/mol = 2.17e-6 mol

n(K) = 0.00005 g / 39 g/mol = 1.28e-6 mol

So the concentration will be:

C(Na) = 2.17e-6 mol / 0.01 L = 0.000217 mol/L

C(K) = 1.28e-6 mol / 0.01 L = 0.000128mol//L

Answer: 0.000217mol/L, 0.000128mol/L

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