A solution of sodium carbonate contains 1.06g of Na2CO3 dissolved in 100ml of water. Calculate its concentration in normality. (d=2.54g/ml)
To calculate the concentration of sodium carbonate in normality, we need the molecular weight of Na2CO3.
The molecular weight of Na2CO3 can be calculated as follows:
Na = 22.99 g/mol (2 Na atoms)
C = 12.01 g/mol
O = 16.00 g/mol (3 oxygen atoms)
Total molecular weight = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
Next, we need to calculate the number of moles of Na2CO3 in the given mass.
1.06 g Na2CO3 × (1 mol Na2CO3 / 105.99 g Na2CO3) = 0.00999 mol Na2CO3
Since normality is defined as the number of equivalents of a solute per liter of solution, we need to determine the number of equivalents of Na2CO3.
Na2CO3 contains two equivalents of sodium ions (Na+) per mole. Therefore, the number of equivalents of Na2CO3 is twice the number of moles.
Number of equivalents = 0.00999 mol Na2CO3 × 2 equivalents/mol = 0.01998 equivalents
Now we can calculate the normality.
Normality = Number of equivalents / Volume of solution in liters
We were given that the volume of the solution is 100 ml, which is equivalent to 0.1 liters.
Normality = 0.01998 equivalents / 0.1 L = 0.1998 N
Therefore, the concentration of the sodium carbonate solution is 0.1998 normality.
To calculate the concentration of the sodium carbonate solution in normality, we need to determine the number of equivalent moles of sodium carbonate present in the solution.
First, let's calculate the number of moles of Na2CO3 using the given mass.
1. Calculate the number of moles:
Molar mass of Na2CO3 = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
= (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.0 g/mol)
= 46.0 g/mol + 12.01 g/mol + 48.0 g/mol
= 106.01 g/mol
Number of moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
= 1.06 g / 106.01 g/mol
= 0.010 moles
Next, let's calculate the volume of the solution in liters.
The volume given in the question is 100 mL. To convert it to liters, divide by 1000.
Volume of solution = 100 mL / 1000
= 0.1 L
Finally, we can calculate the concentration in normality.
Normality (N) = moles of solute / volume of solution in liters
= 0.010 moles / 0.1 L
= 0.1 N
Therefore, the concentration of the sodium carbonate solution in normality is 0.1 N.
Normality= molarity*H+ equivalents= mass/(formulamass)*1/liters*H+
= 1/106 * 1/.1 * 2= .189N