Question

How many grams of sodium carbonate are needed to make 1.50 L of 0.880 M sodium carbonate solution

Answer 1

Answer: 139.9 g/mol of Na2CO3

Step-by-step explanation:

Molarity is equal to the number of moles divided by the Volume. Its expressed as M = #mol / V. You would then change the equation by multiplying both sides by the volume to get 0.880 x 1.5 = 1.32 moles. This number is divided multiplied by 105.98 which is the molar mall of sodium carbonate to get 139.8936 or 139.9 grams

Answer 2

To make a 1.50 L solution of 0.880 M sodium carbonate, one would need 139.91 grams of sodium carbonate, calculated by multiplying the number of moles (1.32 mol) by the molar mass of sodium carbonate (105.99 g/mol).

Step-by-step explanation:

To calculate how many grams of sodium carbonate are needed to make a 1.50 L solution with a concentration of 0.880 M (molar), you first need to determine the number of moles of sodium carbonate required. This is done by multiplying the volume of the solution by the molarity:

Number of moles = Volume (L) × Molarity (M)

1.50 L × 0.880 M = 1.32 moles

Next, you multiply the number of moles by the molar mass of sodium carbonate (Na2CO3) to find the mass needed:

Molar mass of Na2CO3 ≈ 105.99 g/mol

Mass of sodium carbonate = Number of moles × Molar mass

1.32 moles × 105.99 g/mol = 139.91 grams

Therefore, you will need 139.91 grams of sodium carbonate to make a 1.50 L solution of 0.880 M sodium carbonate.