Find the number of mg Na2co3 FW = 106g/mol required to prepare 500ml of 9.2ppm NA solution.
ppm = mg/L
9.2mg/1000ml
x= 0.0092 x 500ml = 4.6mg x 23 = 106mg
The right answer 10.6mg
NA weight = 23
Who helps me to solve it for me.
1 ppm = 1 mg/L
9.2 ppm = 9.2 ppm x 0.500L = 4.6 mg Na.
Convert that to mg Na2CO3 by
4.6 mg x (106/23) = ? mg Na2CO3. Dissolve in water and make up to 0.500 L.
To find the number of mg of Na2CO3 required to prepare a 500ml solution with a concentration of 9.2ppm Na, you need to follow these steps:
Step 1: Determine the concentration of Na in mg/L.
Given that 1 ppm = 1 mg/L, we can convert the concentration of 9.2 ppm to mg/L:
9.2 ppm = 9.2 mg/L
Step 2: Calculate the amount of Na in mg for the given volume.
The concentration is given in mg/L, so to find the amount of Na in a 500ml solution, we multiply the concentration (9.2 mg/L) by the volume in liters:
9.2 mg/L x (500ml / 1000) = 4.6 mg
Step 3: Convert the amount of Na to the molecular weight of Na2CO3.
To convert the amount of Na (in mg) to the amount of Na2CO3 required, we need to consider that the molecular weight of Na2CO3 is 106 g/mol. Since 1 mole = 1000 mg, we can use the formula:
Amount (in mg) = Amount (in moles) x Molecular Weight (in g/mol)
4.6 mg = Amount (in moles) x 106 g/mol
Rearranging the equation to solve for "Amount (in moles)", we have:
Amount (in moles) = 4.6 mg / 106 g/mol
Step 4: Convert the amount of Na2CO3 to mg.
To find the number of mg of Na2CO3 required, we multiply the amount in moles by the molecular weight of Na2CO3:
Amount (in mg) = Amount (in moles) x Molecular Weight (in g/mol)
Amount (in mg) = (4.6 mg / 106 g/mol) x 106 g/mol
Amount (in mg) = 4.6 mg
Therefore, the correct answer is 4.6 mg, not 10.6 mg as you mentioned.