To find the mass of anhydrous Sodium trioxocarbonate(vi) present in 750 cm³ of a 0.1 mol/dm³ solution, we can use the formula:
Mass = Volume x Concentration x Molar mass
First, we need to convert the volume from cm³ to dm³. Since 1 dm³ = 1000 cm³, 750 cm³ is equal to 0.75 dm³.
Next, we need to calculate the number of moles of Sodium trioxocarbonate(vi) using the concentration and volume. The formula for moles is:
Moles = Concentration x Volume
Moles = 0.1 mol/dm³ x 0.75 dm³ = 0.075 mol
Finally, we can calculate the mass using the molar mass of Sodium trioxocarbonate(vi), which is 105.99 g/mol.
Mass = Moles x Molar mass
Mass = 0.075 mol x 105.99 g/mol
Mass = 7.94925 g
Therefore, the mass of anhydrous Sodium trioxocarbonate(vi) present in 750 cm³ of a 0.1 mol/dm³ solution is approximately 7.95 grams.
What mass of anhydrous Sodium trioxocarbonate(vi) present in 750cm of 0.1mol/dm
1 answer