The mass of sodium carbonate required to prepare 250ml of 0.4N solution is
Technically, one can't calculate this BECAUSE there is no reaction given; however, we usually assume, when the reaction isn't given, to use total + or - charge. That is 2 in this case; therefore,
mL x N x mew = grams where mew is milliequivalent weight.
equivalent weight then is molar mass/2 = 106/2 = 53 and mew is 0.053.Substitute mL, N, mew and solve for grams.
To determine the mass of sodium carbonate required to prepare a 250ml 0.4N solution, we need to follow these steps:
Step 1: Convert the volume of the solution from milliliters (ml) to liters (L).
Given that the volume is 250ml, we divide by 1000 to convert it to liters:
250ml ÷ 1000 = 0.25L
Step 2: Calculate the number of moles of sodium carbonate (Na2CO3) using the formula:
moles = normality × volume (in liters)
Given that the normality (N) is 0.4N and the volume is 0.25L, we have:
moles = 0.4N × 0.25L = 0.1 moles
Step 3: Find the molar mass of sodium carbonate (Na2CO3).
The molar mass of Na2CO3 is calculated by adding the atomic masses of sodium (Na), carbon (C), and oxygen (O):
Na: 2 atoms × 22.99 g/mol = 45.98 g/mol
C: 1 atom × 12.01 g/mol = 12.01 g/mol
O: 3 atoms × 16.00 g/mol = 48.00 g/mol
Total molar mass = 45.98 g/mol + 12.01 g/mol + 48.00 g/mol = 105.99 g/mol
Step 4: Calculate the mass of sodium carbonate.
mass = moles × molar mass
mass = 0.1 moles × 105.99 g/mol = 10.599 g
Therefore, the mass of sodium carbonate required to prepare 250ml of 0.4N solution is approximately 10.599 grams.