What mass of anhydrous sodium trioxocarbonate(iv),Na2CO3 is present in 500cm³ of 0.1moldm-³.(Na=23, C=12, O=16).
Volume given=500cm³
Molarity=0.1dm³
Mole= Molarity x volume(cm³)/ 1000
= 0.1 x 500/1000
= 0.05mol
No of moles= Mass/Molar Mass
0.05=m/106
Mass=0.05 x 106
Mass= 5.3g
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Well, I must say, the mass of anhydrous sodium trioxocarbonate(iv), or Na2CO3, is no laughing matter! But fear not, for I, Clown Bot, am here to help.
To find the mass, we need to first calculate the number of moles of Na2CO3 in the solution. We can use the formula:
moles = concentration × volume
Given that the concentration is 0.1 moldm-³ and the volume is 500 cm³ (which we need to convert to dm³), we have:
moles = 0.1 moldm-³ × (500 cm³ ÷ 1000 cm³/dm³)
moles = 0.1 moldm-³ × 0.5 dm³
moles = 0.05 mol
Since the formula of Na2CO3 tells us that there are two moles of sodium (Na) for every one mole of Na2CO3, we multiply the moles by 2:
moles of Na = 0.05 mol × 2
moles of Na = 0.1 mol
Now, using the molar mass of sodium (Na = 23 g/mol), we can find the mass of Na2CO3:
mass = moles × molar mass
mass = 0.1 mol × (23 g/mol + 12 g/mol + (3 × 16 g/mol))
mass = 0.1 mol × (23 g/mol + 12 g/mol + 48 g/mol)
mass = 0.1 mol × 83 g/mol
mass = 8.3 g
So, the mass of anhydrous sodium trioxocarbonate(iv), Na2CO3, present in 500 cm³ of 0.1 moldm-³ solution is approximately 8.3 grams.
To find the mass of anhydrous sodium trioxocarbonate (IV) (Na2CO3) present in 500 cm³ of 0.1 mol/dm³ solution, we can use the formula:
Mass = Concentration * Volume * Molar mass
First, let's calculate the molar mass of Na2CO3. Na has a molar mass of 23 g/mol, C has a molar mass of 12 g/mol, and O has a molar mass of 16 g/mol. Since there are two Na atoms in Na2CO3, we multiply the molar mass of Na by 2:
Molar mass of Na2CO3 = (2 * 23) + 12 + (3 * 16) = 46 + 12 + 48 = 106 g/mol
Next, we can use the given concentration and volume to find the number of moles of Na2CO3 in the solution:
Concentration = 0.1 mol/dm³
Volume = 500 cm³ = 0.5 dm³
Number of moles = Concentration * Volume = 0.1 mol/dm³ * 0.5 dm³ = 0.05 mol
Finally, we can calculate the mass of Na2CO3 using the formula:
Mass = Number of moles * Molar mass = 0.05 mol * 106 g/mol = 5.3 g
Therefore, the mass of anhydrous sodium trioxocarbonate (IV) present in 500 cm³ of 0.1 mol/dm³ is 5.3 grams.